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REDOX  REACTIONS

Redox reaction

Redox reactions are reactions where both oxidation and reduction are taking place.

Oxidation is the loss of electrons by a reactant.

When a metal element is reacting to form a compound then it is being oxidised.

For example:

                          Mg(s)+O₂(g)→MgO(s)

The metal atoms are losing electrons to form an ion. They are being oxidised.

                       Mg(s)→Mg₂+(aq)+2e−

This is known as an ion-electron equation.

Reduction is the opposite of oxidation. It is the gain of electrons.

Compounds reacting that result in metal elements being formed are examples of reduction reactions.

For example:

                     Cu²⁺(aq)+2e−→Cu(s)

The metal ions are gaining electrons to form atoms of the element. They are being reduced.

 Redox reactions takes place in the same reaction and at the same time and they result into displacement of one species by another

For example, if magnesium was added to copper sulphate solution, the magnesium metal would be oxidised, while the copper ions were being reduced.

                   Cu(s)→Cu²⁺(aq)+2e− oxidation reaction

                   2Ag+(aq)+2e−→2Ag(s) reduction reaction

                 Cu(s)+2Ag+(aq)→Cu₂+(aq)+2Ag(s) redox reaction

The superscript along with the sign is, called ‘oxidation state’ of the atom.

The oxidation number

The oxidation number of an atom is the charge it appears to have when you count the electrons according to some arbitrary rules.

By definition, the oxidation number of an atom is the charge that atom would have if the compound was composed of ions.

It is often useful to follow chemical reactions by looking at changes in the oxidation numbers of the atoms in each compound during the reaction. Oxidation numbers also play an important role in the systematic nomenclature of chemical compounds.

1. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in O₂, O₃, P₄, S₈, and aluminum metal all have an oxidation number of 0.
2. The oxidation number of simple ions is equal to the charge on the ion. The oxidation number of sodium in the Na⁺ion is +1, for example, and the oxidation number of chlorine in the Cl^–ion is -1.
3. The oxidation number of hydrogen is +1 when it is combined with a nonmetalas in CH₄, NH₃, H₂O, and HCl.
4. The oxidation number of hydrogen is -1 when it is combined with a metalas in. LiH, NaH, CaH₂, and LiAlH₄.
5. The metals in Group IA form compounds (such as Li₃N and Na₂S) in which the metal atom has an oxidation number of +1.
6. The elements in Group IIA form compounds (such as Mg₃N₂and CaCO₃) in which the metal atom has a +2 oxidation number.
7. Oxygen usually has an oxidation number of -2. Exceptions include molecules and polyatomic ions that contain O-O bonds, such as O₂, O₃, H₂O₂, and the O₂^2-ion.
8. The elements in Group VIIA often form compounds (such as AlF₃, HCl, and ZnBr₂) in which the nonmetal has a -1 oxidation number.
9. The sum of the oxidation numbers in a neutral compound is zero.

H₂O: 2(+1) + (-2) = 0

10. The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion. The oxidation number of the sulfur atom in the SO₄^2-ion must be +6, for example, because the sum of the oxidation numbers of the atoms in this ion must equal -2.

SO₄^2-: (+6) + 4(-2) = -2

11. Elements toward the bottom left corner of the periodic table are more likely to have positive oxidation numbers than those toward the upper right corner of the table. Sulfur has a positive oxidation number in SO₂, for example, because it is below oxygen in the periodic table.

SO₂: (+4) + 2(-2) = 0

In redox reactions:

In redox reactions, atoms or ions either loss or gain electrons and have different oxidation states, before and after the reaction.

• Oxidation number can be positive or zero or negative
• Oxidation number has to be an integer as the number of electrons can only be an integer.
• Oxidation number cannot be fractional
• The oxidation number is the same as the oxidation state.

Balancing Redox Reactions

Oxidation-Reduction or “redox” reactions occur when elements in a chemical reaction gain or lose electrons, causing an increase or decrease in oxidation numbers. The Half Equation Method is used to balance these reactions.

In a redox reaction, one or more element becomes oxidized, and one or more element becomes reduced. Oxidation is the loss of electrons whereas reduction is the gain of electrons. An easy way to remember this is to think of the charges: an element’s charge is reduced if it gains electrons (an acronym to remember the difference is LEO = Lose Electron Oxidation & GER = Gain Electron Reduction). Redox reactions usually occur in one of two environments: acidic or basic. In order to balance redox equations, understanding oxidation states is necessary.

Some points to remember when balancing redox reactions

• The equation is separated into two half-equations, one for oxidation, and one for reduction.
• The equation is balanced by adjusting coefficients and adding H₂O, H⁺, and e^–in this order:
  1. Balance the atoms in the equation, apart from O and H.
  2. To balance the Oxygen atoms, add the appropriate number of water (H₂O) molecules to the other side.
  3. To balance the Hydrogen atoms (including those added in step 2), add H⁺
  4. Add up the charges on each side. They must be made equal by adding enough electrons (e^–) to the more positive side.
• The e^–on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers to be made the same.
• The half-equations are added together, cancelling out the electrons to form one balanced equation. Cancel out as much as possible.

These reactions can take place in either acidic or basic solutions.

• (If the equation is being balanced in a basic solution, the appropriate number of OH^– must be added to turn the remaining H⁺ into water molecules)
• The equation can now be checked to make sure it is balanced.

1. In Acidic Aqueous Solution

Balance this reaction

               MnO⁻₄+I⁻⟶I₂+Mn2⁺

Solution

Steps to balance:

Step 1: Separate the half-reactions that undergo oxidation and reduction.

Oxidation:

                I⁻⟶I₂

This is the oxidation half because the oxidation state changes from -1 on the left side to 0 on the right side. This indicates a gain in electrons.

Reduction:

                MnO⁻₄⟶Mn²⁺

This is the reduction half because the oxidation state changes from +7 on the left side to +2 on the right side. This indicates a reduction in electrons.

Step 2: In order to balance this half reaction we must start by balancing all atoms other than any Hydrogen or Oxygen atoms.

Oxidation:

               2I⁻⟶I₂

In order to balance the oxidation half of the reaction you must first add a 2 in front of the II on the left hand side so there is an equal number of atoms on both sides.

Reduction:

            MnO⁻₄⟶Mn²⁺

For the reduction half of the reaction, you can notice that all atoms other than Hydrogen and Oxygen are already balanced because there is one manganese atom on both sides of the half reaction.

Step 3: Balance Oxygen atoms by adding H₂O to the side of the equation that needs Oxygen. Once you have completed this step add H+ to the side of the equation that lacks H atoms necessary to be balanced.

Oxidation:

                      2I⁻⟶I²

Because there are no Oxygen or Hydrogen atoms in this half of the reaction, it is not required to perform any kind of balancing.

Reduction:

                 MnO⁻⁴⟶Mn²⁺ +  4H₂O

The first step in balancing this reaction using step 3 is to add4 H₂O atoms in order to balance the Oxygen atoms with the 4 on the other side of MnO₄^–

 

Reduction:

                 MnO⁻₄+8H⁺⟶Mn²⁺+4H₂O

Now that the Oxygen atoms have been balanced you can see that there are 8 H atoms on the right hand side of the equation and none on the left. Therefore, you must add 8 H⁺ atoms to the left hand side of the equation to make it balanced.

Step 4: Now that the two half reactions have been balanced correctly one must balance the charges in each half reaction so that both the reduction and oxidation halves of the reaction consume the same number of electrons.

Oxidation:

           2I⁻⟶I₂+2e⁻

Because of the fact that there are two I’s on the left hand side of the equation which a charge of -1 we can state that the left hand side has an overall charge of -2. The I on the left side of the equation has an overall charge of 0. Therefore to balance the charges of this reaction we must add 2 electrons to the right side of the equation so that both sides of the equation have equal charges of -2.

Reduction:

           5e⁻+8H⁺+MnO⁻₄⟶Mn²⁺+4H₂O

Looking at the left hand side of the equation you can notice that there are 8 Hydrogen atoms with a +1 charge. There is also a MnO⁻₄ ion that has a charge of -1. When we add these two charges up we can calculate that the left hand side of the equation has an overall charge of +7. The right hand side has an Mn atom with a charge of +2 and then 4 water molecules that have charges of 0. Therefore, the overall charge of the right side is +2. We must add 5 electrons to the left side of the equation to make sure that both sides of the equation have equal charges of +2.

Step 5: Multiply both sides of both reactions by the least common multiple that will allow the half-reactions to have the same number of electrons and cancel each other out.

Oxidation:

                    10I⁻→5I₂+10e⁻

We multiply this half reaction by 5 to come up with the following result above.

Reduction: 10e−+16H⁺+2MnO⁻₄→2Mn²⁺+8H₂O  

We multiply the reduction half of the reaction by 2 and arrive at the answer above.

By multiplying the oxidation half by 5 and the reduction half by 2 we are able to observe that both half-reactions have 10 electrons and are therefore are able to cancel each other out.

Step 6: Add the two half reactions in order to obtain the overall equation by canceling out the electrons and any H₂O and H⁺ ions that exist on both sides of the equation.

Overall:

          10I⁻+16H⁺+2MnO⁻₄⟶5I₂+2Mn²⁺+8H₂O

In this problem, there is not anything that exists on both halves of the equation that can be cancelled out other than the electrons. Finally, double check your work to make sure that the mass and charge are both balanced. To double check this equation you can notice that everything is balanced because both sides of the equation have an overall charge of +4.

1. In Basic Aqueous Solutions

The balancing procedure in basic solution differs slightly because OH⁻ ions must be used instead of H⁺ ions when balancing hydrogen atoms. To give the previous reaction under basic conditions, sixteen OH⁻ ions can be added to both sides. on the left the OH⁻ and the H⁺ ions will react to form water, which will cancel out with some of the H²Oon the right.

10I⁻(aq)+2MnO⁻₄(aq)+16H⁺(aq)+16OH⁻(aq)⟶5I₂(s)+2Mn²⁺(aq)+8H₂O(l)+16OH⁻(aq)

On the left side the OH- and the H+ ions will react to form water, which will cancel out with some of the H2O on the right:

10I⁻(aq)+2MnO⁻₄(aq)+16H₂O(l)⟶5I₂(s)+2Mn²⁺(aq)+8H₂O(l)+16OH⁻ (aq)

Eight water molecules can be canceled, leaving eight on the reactant side:

10I⁻⁽aq)+2MnO⁻₄(aq)+8H₂O(l)⟶5I₂(s)+2Mn²⁺(aq)+16OH⁻(aq)

This is the balanced reaction in basic solution.

Balance the following in an acidic solution.

             SO₃²⁻(aq)+MnO⁻₄(aq)→SO₂−4(aq)+Mn²⁺(aq)

Solution

To balance a redox reaction, first take an equation and separate into two half reaction equations specifically oxidation and reduction, and balance them.

Step 1: Split into two half reaction equations: Oxidation and Reduction

• Oxidation: 

                   SO₃²⁻ (aq)⟶SO₄²⁻ (aq)

[ oxidation because oxidation state of sulfur increase from +4 to +6]

• Reduction:

             MnO⁺4(aq)⟶Mn²⁺(aq)

[ Reduction because oxidation state of Mn decreases from +7 to +2]

Step 2: Balance each of the half equations in this order:

• Atoms other than H and O
• O atoms by adding H₂Omolecules with proper coefficients
• H atoms by adding H⁺with proper coefficients

The S and Mn atoms are already balanced,

Balancing O atoms

Oxidation:SO3²⁻ (aq)+H₂O(l)→SO₄²⁻ (aq)

Reduction:MnO₄₍⁻aq→Mn²⁺(aq)+4H₂O(l)

Then balance out H atoms on each side

Oxidation:SO²⁻3(aq)+H₂O(l)→SO₄²⁻ (aq)+2H⁺(aq)

Reduction:MnO4⁻ (aq)+8H⁺→Mn²⁺(aq)+4H₂O(l)

Step 3: Balance the charges of the half reactions by adding electrons

Oxidation:SO₃²⁻ (aq)+H₂O(l)→SO₄²⁻ (aq)+2H⁺(aq)+2e−

Reduction:MnO4⁻ (aq)+8H⁺+5e−→Mn²⁺(aq)+4H₂O(l)

Step 4: Obtain the overall redox equation by combining the half reaction, but multiply entire equation by number of electrons in oxidation with reduction equation, and number of electrons in reduction with oxidation equation.

Oxidation:5×[SO₃²⁻ (aq)+H₂O(l)→SO₄²⁻ (aq)+2H⁺(aq)+2e−]

Reduction:2×[MnO₄⁻ (aq)+8H⁺+5e−→Mn²⁺(aq)+4H₂O(l)]

Overall Reaction:

Oxidation:5SO₃²⁻ (aq)+5H₂O(l)→5SO₄²⁻ (aq)+10H⁺(aq)+10e−

Reduction:2MnO₄⁻ (aq)+16H⁺+10e−→2Mn²⁺(aq)+8H₂O(l)

total:5SO₃²⁻ (aq)+5H₂O(l)+2MnO4⁻ (aq)+16H⁺+10e−→5SO₄²⁻ (aq)+10H⁺(aq)+2Mn²⁺(aq)+8H₂O(l)+10e−

Step 5: Simplify and cancel out similar terms on both sides

To get

5SO₃²⁻(aq)+2MnO4⁻(aq)+6H⁺→5SO₄²⁻(aq)+2Mn²⁺(aq)+3H²O(l)

Balance this reaction in both acidic and basic aqueous solutions

         MnO4⁻(aq)+SO₃²⁻ (aq)⟶MnO2 (s) +SO₄²⁻ (aq)

Solution

First, they are separated into the half-equations:

            MnO₄⁻ (aq)⟶MnO2(s)

This is the reduction half-reaction because oxygen is LOST)

and

            SO₃²⁻ (aq)⟶SO₄²⁻ (aq))

(the oxidation, because oxygen is GAINED)

Now, to balance the oxygen atoms, we must add two water molecules to the right side of the first equation, and one water molecule to the left side of the second equation:

           MnO₄⁻ (aq)⟶MnO₂(s)+2H₂O(l)

           H2O(l)+SO₃²⁻ (aq)⟶SO₄²⁻ (aq)

To balance the hydrogen atoms (those the original equation as well as those added in the last step), we must add four H⁺ ions to the left side of the first equation, and two H⁺ ions to the right side of the second equation.

            4H⁺+MnO₄⁻ (aq)⟶MnO₂(s)+2H₂O(l)

            H₂O(l)+SO₃²⁻ (aq)⟶SO₄²⁻ (aq)+2H⁺

Now we must balance the charges. In the first equation, the charge is +3 on the left and 0 on the right, so we must add three electrons to the left side to make the charges the same. In the second equation, the charge is -2 on the left and 0 on the right, so we must add two electrons to the right.

             3e−+4H⁺+MnO₄⁻ (aq)⟶MnO₂(s)+2H₂O(l)

               H₂O(l)+SO₃²⁻ (aq)−⟶SO₄²⁻ (aq)+2H⁺+2e−

Now we must make the electrons equal each other, so we multiply each equation by the appropriate number to get the common multiple (in this case, by 2 for the first equation, and by 3 for the second).

                2(3e−+4H⁺+MnO₄⁻ (aq)⟶MnO₂(s)+2H₂O(l))

                 3(H₂O(l)+SO₃²⁻ (aq)⟶SO₄²⁻ (aq)+2H⁺+2e−)

With the result:

6e−+8H⁺+2MnO₄⁻(aq)⟶2MnO₂(s)+4H₂O(l)

3H₂O(l)+3SO₃²⁻ (aq)⟶3SO₄²⁻ (aq)+6H⁺+6e−

Now we cancel and add the equations together. We can cancel the 6e^– because they are on both sides. We can get rid of the 6H⁺ on both sides as well, turning the 8H⁺ in the first equation to 2H+. The same method gets rid of the 3H2O(l) on the bottom, leaving us with just one H2O(l) on the top. In the end, the overall reaction should have no electrons remaining. Now we can write one balanced equation:

2MnO₄⁻ (aq)+2H⁺+3SO₃²⁻ (aq)⟶H2O(l)+2MnO₂(s)+3SO₄²⁻ (aq)

The equation is now balanced in an acidic environment.

To balance in a basic environment add OH− to each side to neutralize the H+ into water molecules:

2MnO₄⁻ (aq)+2H₂O+3SO₃²⁻ (aq)⟶H₂O(l)+2MnO₂(s)+3SO₄²⁻ (aq)+2OH−

and then cancel the water molecules

2MnO₄⁻(aq)+H₂O+3SO₃²⁻(aq)⟶+2MnO₂(s)+3SO₄²⁻ (aq)+2OH−

The equation is now balanced in a basic environment.

Balance this reaction in acidic solution

Fe(OH)³⁺OCl−→FeO₄²⁻+Cl⁻

 

Solution

Step 1:

• Reduction:

                       OCl⁻⟶Cl⁻

• Oxidation:

            Fe(OH)₃⟶FeO₄ ²⁻ 

Steps 2 and 3:

• Reduction:

2H⁺+OCl⁻+2e−⟶Cl⁻+H₂O

• Oxidation:

Fe(OH)₃+H₂O⟶FeO₄²⁻+3e−+5H⁺ 

Overall Equation:

3×[2H⁺+OCl⁻+2e−⟶Cl⁻+H₂O]

6H⁺+3OCl⁻+6e−⟶3Cl⁻+3H₂O

and

2×[Fe(OH)₃⁺H₂O ⟶FeO4²⁻+3e⁻+5H⁺]⟶2FeO₄⁻²+6e−+10H⁺

Adding these together results

6H⁺+3OCl⁻+2e−+2Fe(OH)₃+2H₂O⟶3Cl⁻+3H₂O+2FeO₄²⁻+6e−+10H+

Step 5:

Simplify:

3OCl⁻+2Fe(OH)₃→3Cl₋+H₂O+2FeO₄²⁻+4H⁺

Balance this equation in acidic aqueous solution

VO₄³⁻+Fe²⁺⟶VO²⁺+Fe³⁺

Step 1:

• Oxidation:

                   Fe²⁺⟶Fe³⁺

• Reduction:

                  VO₄³⁻⟶VO²⁺

Step 2/3:

• Oxidation:

                    Fe²⁺⟶Fe³⁺+e−

 Reduction:

                      6H⁺+VO₄³⁻+e−⟶VO²⁺+3H₂O

Step 4:

Overall Reaction:

                         Fe²⁺⟶Fe³⁺+e−

                     6H⁺+VO₄³⁻+e−⟶VO²⁺+3H₂O

             Fe²⁺+6H⁺+VO₄³⁻+e−→Fe³⁺+e−+VO²⁺+3H₂O

Step 5:

Simplify:

                   Fe²⁺+6H⁺+VO₄³⁻→Fe³⁺+VO²⁺+3H₂O

5. calculate the Oxidation number of Manganese in permanganate ion MnO₄^–

Charge on the permanganate ion is -1

Oxidation state of permanganate ion =Oxidation state of manganese + 4 oxidation state of oxygen = -1.

Oxidation state of oxygen = -2

Oxidation states → x + (4*-2) = -1: x = +7

Atoms in the species → Mn 4O

Oxidation state of manganese = +7

2. Work  out the oxidation numbers of   the following metal ion in a complex.
3. i) Ni(CO)₄.

The total charge of the complex is zero. CO is a neutral molecule.

Oxidation states → x + (4*0) = 0: x = 0

Atoms in the species → Ni 4 CO

Nickel is also in zero oxidation state.

1. ii) [CoCl₂(NH₃)₄]Cl.

The complex can be written in the ionic forms as [CoCl₂(NH₃)₄]⁺Cl^–.

Metal is in a cationic complex with a unitary positive charge. Ammonia is a neutral ligand and chlorine has a unit negative charge.

Oxidation number of [CoCl₂(NH₃)₄]⁺ 

= Oxidation number of (Co + 2Cl + 4×0) = +1.

Oxidation states → x + (2*-1) + 4*0 = +1: x = +3

Atoms in the species → Co 2Cl 4 NH₃

Oxidation number of cobalt in the complex = +3

3. Calculate the  oxidation numbers in  Cl₂O, Cl₂O₅and Cl₂O₇.
4. i) Cl₂O:

Cl₂O is neutral and so, net charge=0.

Net oxidation state of Cl₂O = 2 x

Oxidation state of chlorine + 1x

Oxidation state of oxygen = -2.

Oxidation states → 2 x + (-2) = 0:

                                 2x= 2

 x = +1

Cl= 1

 

1. ii) Cl₂O₅:

Cl₂O₅ is neutral and so, net charge=0

Oxidation state of chlorine + 5 x oxidation state of oxygen = 0.

⸪ Oxidation state of oxygen = -2.

Oxidation states → 2x + (5 X -2) = 0:

                               2x = 10

x = +5

Oxidation state of chlorine in Cl₂O₅  = +5

 

iii) Cl₂O₇:

Cl₂O₇ is neutral and so, net charge=0

Oxidation state of Cl₂O₇ = 2 X Oxidation state of chlorine + 7 X oxidation state of oxygen = 0.

⸪ Oxidation state of oxygen = -2.

Oxidation states → 2x + (7X-2) = 0:

                                2x= 14

x = +7

Oxidation state of chlorine in Cl₂O = +7

Note: Except the atoms/molecules/ions mentioned, as having a constant oxidation state, oxidation state of other atoms/molecule and ions will vary depending on the molecule they are present.

In the given examples, the oxidation state of chlorine is not constant, but variable (+1, +5 and +7)

6.Calculate the Oxidation state of chromium in dichromate anion.

Dichromate ion is Cr₂O₇^2-.

Charge on the ion is -2.

Oxidation state of dichromate ion

= 2  X (Oxidation state of chromium )+ 7 X (oxidation state of oxygen) = -2.

Oxidation state of oxygen = -2.

Oxidation states → 2x + (7X-2) = -2:

                                    2x= -2+14

                                          2x= 12

                                           x = +6

Oxidation state of chromium= 12 / 2 = 6

8. Calculate the oxidation  state of Chlorine in Cl₂O₄
9. i) The average oxidation state of chlorine

Cl₂O₄ is neutral and so, net charge = 0

Oxidation state of Cl₂O₄ = 2 x( Oxidation state of chlorine) + 4 x (oxidation state of oxygen) = 0.

⸪ Oxidation states → 2x + (4X-2) = 0:

                                  2x= 8

                                      x = +4

Oxidation state of chlorine in Cl₂O₅ =  +4

 

11.Given the following two half–reactions:

(i) MnO₄^–_(aq) + 8H⁺_(aq)  + 5e^–==> Mn²⁺_(aq) + 4H₂O_(l)

and (ii) Fe³⁺_(aq) +  e^– ==> Fe²⁺_(aq)

• Construct the fully balanced redox ionic equationfor the manganate(VII) ion oxidising the iron(II) ion:

 MnO₄^–_(aq) + 8H⁺_(aq)  + 5Fe²⁺_(aq) ===> Mn²⁺_(aq) + 5Fe³⁺_(aq) + 4H₂O_(l) 

(b) 24.3 cm³ of 0.0200 mol dm^–3 KMnO₄ reacted with 20.0 cm³ of an iron(II) solution acidified with dilute sulfuric acid.

(i) Calculate the molarity of the iron(II) ion.

mol MnO₄^– = 0.0200 x 24.3 / 1000

                  = 0.000486

therefore mol Fe²⁺ = 5 x 0.000486 (1 : 5 in equation)

                            = 0.00243 in 20 cm³

so scaling up to 1 dm³, the molarity of  Fe²⁺ 

         = 0.00243 x 1000 / 20

               = 0.122 mol dm^–3.

(ii) How do recognise the end–point in the titration?

The end point is the first faint permanent pink due to a trace excess of KMnO₄.

 

1. Distinguish between oxidation and reduction  in terms of

• Electron transfer
• Oxidation number

12.Calculate the percentage of iron in a sample of steel wire if 1.51 g of the wire was dissolved in excess of dilute sulfuric acid and the solution made up to 250 cm³ in a standard graduated flask. A 25.0 cm³ aliquot of this solution was pipetted into a conical flask and needed 25.45 cm³ of  O.0200 mol dm^–3 KMnO₄ for complete oxidation.

mol MnO₄^– = 0.0200 x 25.45 / 1000 = 0.000509

mol Fe = 5 x 0.000509 = 0.002545,

mass Fe = 0.002545 x 55.9 = 0.1423 g

total Fe in wire = 0.1423 x 10 = 1.423 g (only ¹/₁₀th of the made up solution used in titration)

so % Fe = 1.423 x 100 / 1.51

              = 94.2 %

(a) Suggest reasons why the presence of dil. sulfuric acid is essential for an accurate titration and why dil. hydrochloric and nitric acids are unsuitable to be used in this context.

The titrations are done with (i) dilute sulfuric acid present to prevent side reactions e.g. MnO₂ formation (brown colouration or black precipitate).

(ii) Nitric acid is an oxidising agent and would oxidise iron(II) ions to iron(III) ions,

and (iii) chloride ions are oxidised to chlorine by manganate(VII) ions (see half-cell potential data in Q3c),

therefore, using either acid (ii) or (iii) would give a false titration result.

1. The analysis of a soluble iron(II) salt to obtain the percentage of iron in it.25g of an iron(II) salt was dissolved in 250 cm³ of pure water. 25.0 cm³ aliquots were pipetted from this stock solution and titrated with 0.0200 mol dm^–3 potassium manganate(VII) solution.

The titration values obtained were 23.95 cm³, 23.80 cm³ and 23.85 cm³.

(i) What titration value should be used in the calculation and why? 

23.87 cm³, i.e. the average titration value, which is statistically more accurate than an individual titration result.

(ii) molarity = mol / volume in dm³

BUT, remember, 1 dm³ = 1000 cm³,

burettes and pipettes work in cm³ and molarity works in dm³,

a bit of pain I know, but live with it and go with the flow!

therefore mol = molarity x vol in cm³ / 1000

mol MnO₄^– = 0.0200 x 23.87/1000 = 4.774 x 10^–4

From the balanced redox equation, 1 mol MnO₄^– oxidises 5 mol of iron(II) ions

mol Fe²⁺ = 5 x 4.774 x 10^–4 = 2.387 x 10^–3

(ii) Calculate the moles of manganate(VII) used in the titration.

You can assume mass of Fe = mass of Fe²⁺ ion, (2 electrons don’t weigh much!)

mol = mass/atomic mass, so mass = mol x atomic mass

mass Fe = 2.387 x 10^–3 x 55.8 = 0.1332 g

(iii) calculate the moles of iron(II) ion titrated Total mass of iron in original sample = 10 x 0.1332 = 1.332 g (scaling up by factor of 250/25)

(iv) Calculate the mass of iron(II) titrated

You can assume mass of Fe = mass of Fe²⁺ ion, (2 electrons don’t weigh much!)

mol = mass/atomic mass, so mass = mol x atomic mass

mass Fe = 2.387 x 10^–3 x 55.8 = 0.1332 g

(v) Calculate the total mass of iron in the original sample of the iron(II) salt. 

Total mass of iron in original sample

 = 10 x 0.1332 = 1.332 g 

(vi) calculate the % iron in the salt. 

% iron in the original salt = 1.332 x 100 / 8.25 = 16.1% (1dp, 3sf)

2. Given the following two half–reactions
   1. Given
   2. (i) S₄O₆^2–_(aq) + 2e^– ==> 2S₂O₃^2–_(aq) 

and

 (ii) I_2(aq) + 2e^– ==> 2I^–_(aq)

construct the full ionic redox equation for the reaction of the thiosulfate ion S₂O₃^2– and iodine I₂.

(i)  I₂(aq) + 2e^– ==> 2I^–(aq)   (iodine reduced by thiosulfate ion, reduction is electron gain)

(ii) 2S₂O₃^2–(aq)  ==>  S₄O₆^2–(aq) + 2e^–  (thiosulfate ion oxidised by iodine, oxidation is electron loss)

(Note that you needed to reverse the half-cell equation as given in question to arrive at the full ionic redox equation)

(i) + (ii) gives: 2S₂O₃^2–(aq)  +  I₂(aq)  ==>  S₄O₆^2–(aq) + 2I^–(aq)

(b) what mass of iodine reacts with 23.5 cm³ of 0.0120 mol dm^–3 sodium thiosulfate solution.

mol S₂O₃^2– = 0.0120 x 23.5 / 1000

             = 0.000282, mole iodine as I₂

            = mol  S₂O₃^2– / 2 (1 : 2 in equation)

              = 0.000141,

A_r(I) = 126.9, so M_r(I₂)

            = 2 x 126.9

            = 253.8

therefore: mass of iodine = 0.000141 x 126.9 x 2 = 0.0358 g

3. 0 cm³ of a solution of iodine in potassium iodide solution required 26.5 cm³ of 0.0950 mol dm^–3 sodium thiosulfate solution to titrate the iodine.

What is the molarity of the iodine solution and the mass of iodine per dm³?

mol S₂O₃^2– = 0.095 x 26.5 / 1000

               = 0.002518,

mol of iodine = mole ‘thio’ / 2

          = 0.002518 / 2

             = 0.001259 in 25.0 cm³,

scaling up to 1 dm³ gives 0.001259 x 1000 /25

      = 0.0504 mol dm^–3 of molecular iodine I₂.

mass concentration of I₂ = 0.0504 x 2 x 126.9

                          = 12.8 g dm^–3 of iodine

13. 2.83g of a sample of haematite iron ore [iron (III) oxide, Fe₂O₃] were dissolved in concentrated hydrochloric acid and the solution diluted to 250cm³.

25.0 cm³ of this solution was reduced with tin(II) chloride (which is oxidised to Sn⁴⁺ in the process) to form a solution of iron(II) ions. This solution of iron(II) ions required 26.4 cm³ of a 0.0200 mol dm^–3 potassium dichromate(VI) solution for complete oxidation back to iron(III) ions.

(a) given the half–cell reactions 

(i) Sn⁴⁺_(aq) + 2e^– ==> Sn²⁺_(aq)

And

 (ii) Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6e^– ==> 2Cr³⁺_(aq) + 7H₂O_(l)

deduce the fully balanced redox equations for the reactions

(i) the reduction of iron(III) ions by tin(II) ions Sn²⁺_(aq) + 2Fe³⁺_(aq) ==> Sn⁴⁺_(aq) + 2Fe²⁺_(aq)

(ii) the oxidation of iron(II) ions by the dichromate(VI) ion 

Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) ==> 2Cr³⁺_(aq) + 6Fe³⁺_(aq) + 7H₂O_(l)

• Calculate the percentage of iron(III) oxide in the ore.

for a 25.0 cm³ aliquot titrated

mol Cr₂O₇^2– = 0.0200 x 26.4 / 1000

                   = 0.000528,

mol Fe²⁺ titrated = 6 x Cr₂O₇^2– = 0.003168

(from 1 mol Cr₂O₇^2– : 6 mol Fe²⁺ in the redox equation),

mol Fe₂O₃ = mol Fe²⁺ / 2 = mol Fe³⁺ / 2

                             = 0.003168 / 2

                                =  0.001584

(moles Fe²⁺ or Fe³⁺ halved, because of 2 moles of Fe³⁺ in in one mole of Fe₂O₃)

M_r(Fe₂O₃) = 159.8 (Fe = 55.9, 0 = 16.0)

so mass of Fe₂O₃ = 0.001584 x 159.8 = 0.2531 g.

Total mass of Fe₂O₃ = 0.2531 x 10 = 2.531 g

 Remember only ¹/₁₀th titrated, so need to scale up by 10 to account for all the iron oxide in the original sample.

Therefore % Fe₂O₃ = 2.531 x 100 / 2.83 = 89.4%

Note that that overall the ratio of mol Cr₂O₇^2– : mol Fe₂O₃ is 3 : 1.

Now its ok to solve this problem using the 3 : 1 ratio if you are very confident to use short cuts in this type of calculation. However, I think from a teaching and learning point of view, I prefer to show the full logic of each step in the calculation, particularly when dealing with mole ratios and this is what I’ve tried to do on this page!

• Suggest why potassium manganate(VII) isn’t used for this titration? (though it was ok in Q1)

Potassium manganate(VII) isn’t used for this titration because it is strong enough to oxidise chloride ions (from the hydrochloric acid) to form chlorine, giving a completely false titration.

14. An approximately 02mol dm^–3 potassium manganite (VII) solution was standardized against precisely 0.100 mol dm^–3 iron(II) ammonium sulfate solution. 25.0 cm³ of the solution of the iron(II) salt were oxidized by 24.15 cm³ of the manganate(VII) solution.

What is the molarity of the potassium manganate(VII) solution ?

mol Fe²⁺ = 0.100 x 25.0 / 1000 = 0.0025,

mol MnO₄^– = mol Fe²⁺ / 5 (from equation 1 : 5)

                      = 0.0005 in 24.15 cm³,

scaling up to 1 dm³, molarity of MnO₄^– = 0.0005 x 1000 / 24.15 = 0.0207 mol dm^–3.

15. 0g of iron(II) ammonium sulfate crystals were made up to 250 cm³ of acidified aqueous solution. 25.0 cm³ of this solution required 21.25 cm³ of 0.0200 mol dm^–3 potassium dichromate(VI) for oxidation.

Calculate x in the formula FeSO₄.(NH₄)₂SO₄.xH₂O

 mol Cr₂O₇^2– = 0.0200 x 21.25 / 1000 = 0.000425,

mol of Fe salt = mol Fe²⁺ titrated = 6 x Cr₂O₇^2– 

                                         = 6 x 0.000425 = 0.00255,

BUT only ¹/₁₀th of Fe²⁺ salt used in titration,

so 1 g of FeSO₄.(NH₄)₂S0₄ .xH₂O is equal to 0.00255 mol.

Scaling up to 1 mol gives a molar mass for the salt in g mol^–1 of 1 x 1 /0.00255 = 392.2.

So the formula mass for FeSO₄.(NH₄)₂S0₄.xH₂O is 392.2.

Now the formula mass of FeSO₄.(NH₄)₂S0₄ = 284.1,

    this leaves 392.2 – 284.1

                                     = 108.1 mass units.

M_r(H₂O) = 18, so 108.1 / 18 = 6.005 mol of water, so x = 6 in the salt formula, FeSO₄.(NH₄)₂S0₄.6H₂O.

16. Given the half–reaction C₂O₄^2–_(aq) – 2e^– ==> 2CO_2(g)

or H₂C₂O_4(aq) – 2e^– ==> 2CO_2(g) + 2H⁺_(aq)

• write out the balanced redox equation for manganate(VII) ions oxidising the ethanedioate ion (or ethanedioic acid).

2MnO₄^–_(aq) + 16H⁺_(aq)  + 5C₂O₄^2–_(aq) ==> 2Mn²⁺_(aq)  + 8H₂O_(l) + 10CO_2(g)

or  2MnO₄^–_(aq) + 6H⁺_(aq)  + 5H₂C₂O_4(aq) ==> 2Mn²⁺_(aq)  + 8H₂O_(l) + 10CO_2(g)

(b) 1.520 g of ethanedioic acid crystals, H₂C₂O₄.2H₂O, was made up to 250.0 cm³ of aqueous solution and 25.00 cm³ of this solution needed 24.55 cm³ of a potassium manganate(VII) solution for oxidation.

Calculate the molarity of the manganate(VII) solution and its concentration in g dm^–3.

M_r(H₂C₂O₄.2H₂O) = 126.0

total mol H₂C₂O₄.2H₂O (or C₂O₄^2–)  = 1.52 / 126

                             = 0.0120635

but mol of C₂O₄^2– in titration = 0.0120635/10

          = 0.00120635 (¹/₁₀th used, 25 of 250 cm³),

mol MnO₄^– = mol of C₂O₄^2– / 2.5 (2:5 or 1:2.5 ratio),

mol MnO₄^– = 0.00120634 / 2.5

                   = 0.00048254 (in 24.55 cm³),

scaling up to 1 dm³ the molarity of MnO₄^– 

                  = 0.00048254 x 1000 / 24.55

                  = 0.0196554

= 0.01966 mol dm^-3 (5dp, 4sf) OR 0.0197 mol dm^–3 (4dp, 3sf)

M_r(KMnO₄) = 158.0, so in terms of mass concentration 

          = 0.01965523 x 158 = 3.10555

= 3.106 g dm^-3 (3dp, 4sf) OR 3.11 g dm^–3 (2dp, 3sf)

[NOTE: The titration volume, mass and formula masses are quoted to four significant figures (4sf), so it might be considered legitimate to quote the answer to four significant figures]

 

 

17. A standardization of potassium manganate(VII) solution yielded the following data: 150g of potassium tetraoxalate dihydrate,  KHC₂O₄.H₂C₂O₄.2H₂O needed 23.20 cm³ of the manganate(VII) solution.

What is the molarity of the manganate(VII) solution?

mol KHC₂O₄.H₂C₂O₄.2H₂O (M_r = 254.2)

= 0.150 / 254.2

 = 0.000590087

ratio of tetroxalate to manganate(VII) is 2:2.5 or 1:1.25 (note equiv of 2 C₂O₄^2– in salt),

so mol MnO₄^– in titration = 0.000590087 / 1.25

                        = 0.000472069 in 23.2 cm³,

scaling up to 1 dm³ gives for [MnO₄^–] 

                    = 0.000472069 x 1000 / 23.2

                   = 0.02034781

          = 0.02034 mol dm^-3 (5dp, 4sf) OR 0.0203 mol dm^–3 (4dp, 3sf)

[NOTE: Quoting the concentration to 4dp, 3sf is more appropriate here because the mass is only quoted to 3sf and the titration is only likely to be accurate to the nearest 0.05 cm³]

18. Given the half–cell equation  O_2(g)+2H⁺_(aq)+ 2e^– ==> H₂O_2(aq)

(a) construct the fully balanced redox ionic equation for the oxidation of hydrogen peroxide by potassium manganate(VII) )  2MnO₄^–_(aq) + 6H⁺_(aq)  + 5H₂O_2(aq)  ==> 2Mn²⁺_(aq) + 8H₂O_(l) + 5O_2(g)  

(b) 50.0 cm³ of solution of hydrogen peroxide were diluted to 1.00 dm³ with water. 25.0 cm³ of this solution, when acidified with dilute sulfuric acid, reacted with 20.25 cm³ of 0.0200 mol dm^–3 KMnO₄.

What is the concentration of the original hydrogen peroxide solution in mol dm^–3?

 in titration mol MnO₄^– =  0.0200 x 20.25 / 1000 = 0.000405,

MnO₄^–:H₂O₂ ratio is 2:5 or 1:2.5, so mol H₂O₂ in titration = 0.000405 x 2.5 = 0.0010125,

scaling up for total mol H₂O₂ in diluted solution (of 1 dm³ or 1000 cm³) = 0.0010125 x 1000 / 25.0 = 0.0405 mol,

but in the original 50 cm³ solution,

therefore scaling up to 1 dm³, the original molarity of H₂O₂ is

 0.0405 x 1000 / 50 = 0.810 mol dm^–3.

 

19.  13.2 g of iron(III) alum were dissolved in water and reduced to an iron(II) ion solution by zinc and dilute sulfuric acid. The mixture was filtered and the filtrate and washings made up to 500cm³in a standard volumetric flask. If 20.0 cm³ of this solution required 26.5  cm³ of 0.0100 mol dm^–3 KMnO₄ for oxidation.

• give the ionic equation for the reduction of iron(III) ions by zinc metal.

Zn_(s) + 2Fe³⁺_(aq) ==> Zn²⁺_(aq) +  2Fe²⁺_(aq)

• Calculate the percentage by mass of iron in iron alum.

You need to know the redox equation to get the mole ratio involved in this titration

MnO₄^–_(aq) + 8H⁺_(aq)  + 5Fe²⁺_(aq) ==> Mn²⁺_(aq) + 5Fe³⁺_(aq) + 4H₂O_(l)

Changes in oxidation state: Mn from +7 to +2 and Fe from +2 to +3

Therefore, on electron/oxidation state change, 1 mole of MnO₄^– can oxidise 5 moles of Fe²⁺

Therefore:

mol MnO₄^– in  titration = 0.0100 x 26.5 / 1000 =  0.000265,

mol Fe (Fe²⁺) = mol MnO₄^– x 5 = 0.001325 in 20.0 cm³ of the alum solution,

scaling up gives total mol Fe = 0.001325 x 500 / 20 = 0.033125,

total mass Fe in the 13.2 g of alum = 0.033125 x 55.9 = 1.852,

so % Fe = 1.852 x 100 / 13.2 = 14.0%

17. Calculate the concentration in mol dm^–3and g dm^–3, of a sodium ethanedioate (Na₂C₂O₄) solution, 00 cm³ of which were oxidized in acid solution by 24.50 cm³ of a potassium manganate(VII) solution containing 0.05 mol dm^–3.

mol MnO₄^– in titration = 0.05 x 24.5 / 1000 = 0.001225,

ratio MnO₄^–:Na₂C₂O₄ is 2:5 or 1:2.5,

           so mol Na₂C₂O₄ titrated

                      = 0.001225 x 2.5 = 0.003063 in 5 cm³,

scaling up to 1 dm³, molarity Na₂C₂O₄ = 0.003063 x 1000 / 5                           = 0.613  mol dm^–3

M_r(Na₂C₂O₄) = 134, so concentration = 0.613 x 134 = 82.1 g dm^–3

21. Calculate xin the formula FeSO₄.xH₂O from the following data:  12.18g of iron(II) sulfate crystals were made up to 500 cm³ acidified with sulfuric acid.

25.0 cm³ of this solution required 43.85 cm³ of 0.0100 mol dm^–3 KMnO₄ for complete oxidation.

mol KMnO₄ = 0.0100 x 43.85 / 1000

                   = 0.0004385, mol Fe (Fe²⁺)  = mol KMnO₄ x 5,mol Fe                     = 0.0004385 x 5

= 0.0021925, so mol  FeSO₄.xH₂O is also 0.0021925,

in the titration ¹/₂₀th of the salt was used (²⁵/₅₀₀), so ¹/₂₀th of 12.18 g = 0.0021925 mol of the salt = 0.609 g,

scaling up the mass of 1 mole of the salt is 0.609 x 1 / 0.0021925

                                                   = 277.8,

so formula mass of FeSO₄.xH₂O is 277.8, now the formula mass of FeSO₄ is 152.0,

so the formula mass of xH₂O = 277.8 – 152.0 = 125.8,

M_r(H₂O) = 18, so x = 125.8 / 18 = 6.989,

so x = 7 and the formula of the salt is  FeSO₄.7H₂O, i.e. seven molecules of water of crystallisation. 

22.     Given the half–reaction 

                NO₃^–_(aq) + 2H⁺_(aq) + 2e^– ==> NO₂^–_(aq) + H₂O_(l)

• give the ionic equation for potassium manganate(VII) oxidising nitrate(III) to nitrate(V)

2MnO₄^–_(aq) + 6H⁺_(aq)  + 5NO₂^–_(aq) ==> Mn²⁺_(aq) + 5NO₃^–_(aq) +  3H₂O_(l)

(b) 24.2 cm³ of sodium nitrate(III) [sodium nitrite] solution, added from a burette, were needed to discharge the colour of 25.0 cm³ of an acidified 0.0250 mol dm^–3 KMnO₄ solution.

What was the concentration of the nitrate(III)  solution in grammes of anhydrous salt per dm³?

mol KMnO₄ in titration = 0.0250 x 25 / 1000

                                       = 0.000625,

mol ratio MnO₄^–:NO₂^– is 2:5 or 1:2.5, so mol NO₂^– in titration

                      = 0.000625 x 2.5

                        = 0.0015625 in 24.2 cm³,

scaling up to 1 dm³ gives a molar concentration of NaNO₂ of  0.0015625 x 1000 / 24.2

                           = 0.0646 mol dm^–3

M_r(NaNO₂) = 69, so in terms of mass concentration 

                   = 0.0646 x 69

                   = 4.46 g dm^–3

23. 2.68g of iron(II) ethanedioate, FeC₂O₄, were made up to 500cm³ of acidified aqueous solution. 25.0 cm³ of this solution reacted completely with 28.0 cm³ of 0.0200 mol dm^–3 potassium manganate(VII) solution.

Calculate the mole ratio of KMnO₄ to FeC₂O₄ taking part in this reaction. Give the full redox ionic equation for the reaction. M_r(FeC₂O₄) = 143.9, mol FeC₂O₄ in original solution

                  = 2.68 / 143.9 = 0.01862,

scaling down the mol FeC₂O₄ in the titration

                    = 0.01862 x 25 / 500 = 0.000931,

mol KMnO₄ in titration = 0.0200 x 28.0 / 1000

                               = 0.00056,

so ratio KMnO₄:FeC₂O₄ is 0.00056:0.000931 = giving the ‘not so easy to spot’ 3:5 the reacting mole ratio.

FeC₂O₄ is made up of a  Fe²⁺ ion and a  C₂O₄^2– ion, and the full redox equation is:

3MnO₄^–_(aq)+ 5FeC₂O_4(aq) + 24H⁺_(aq)  ==> 3Mn²⁺_(aq) + 5Fe³⁺_(aq)+ 12H₂O_(l) + 10CO_2(g)

or 3MnO₄^–_(aq)+ 5Fe²⁺_(aq) + 5C₂O₄^2–_(aq) + 24H⁺_(aq)  ==> 3Mn²⁺_(aq) + 5Fe³⁺_(aq)+ 12H₂O_(l) + 10CO_2(g)

 

24.    Given the half–cell reaction IO₃^–_(aq)+ 6H⁺_(aq)+ 5e^– ==> ¹/₂I_2(aq) + 3H₂O_(l) (see also Q2)

• Deduce the redox equation for iodate(V) ions oxidising iodide ions.

 IO₃^–_(aq) + 5I^–_(aq) + 6H⁺_(aq) ==> 3I_2(aq) + 3H₂O_(l)

• What volume of 0120mol dm^–3 iodate(V) solution reacts with 20.0 cm³ of 0.100 mol dm^–3 iodide solution?

mol I^– titrated = 0.100 x 20.0 / 1000 = 0.002, mole ratio IO₃^–:I^– is 1:5,

so mole IO₃^– reacted = 0.002 / 5 = 0.0004,

so 0.0004 = 0.012 x (volume IO₃^– required) / 1000,

volume IO₃^– required = 0.0004 x 1000 / 0.012 = 33.3 cm³

(c) 25.0 cm³ of the potassium iodate(V) solution were added to about 15 cm³ of a 15% solution of potassium iodide (ensures excess iodide ion). On acidification, the liberated iodine needed 24.1 cm³ of 0.0500 mol dm^–3 sodium thiosulfate solution to titrate it.

(i) Calculate the concentration of potassium iodate(V) in g dm^–3 

mole S₂O₃^2– (‘thio’) = 0.0500 x 24.1 / 1000

                         = 0.001205,

I₂:S₂O₃^2– ratio is 1:2 in the titration reaction, so mol I₂ 

                      = mole S₂O₃^2– / 2

                      = 0.001205 / 2

                      = 0.0006025,

now the  IO₃^–:I₂ reaction ratio is 1:3,

so mol IO₃^– reacting to give iodine = mole I₂ formed / 3

                 = 0.0006025 / 3 = 0.000201 in 25 cm³,

so scaling up to 1 dm³ the molarity of the KIO₃ (IO₃^–)

                         = 0.000201 x 1000 / 25 

                       = 0.00804 mol dm^–3,

 M_r(KIO₃) = 214.0, so in terms of mass, concentration 

                  = 0.00804 x 214 

                  = 1.72 g dm^–3.

A quicker approach if confident! – ratios from all equations involved are: 2S₂O₃^2– : I₂ : ¹/₃IO₃^–, means that the overall mole iodate(V) = mole thiosulphate / 6, so you can ‘jump’ from line ‘1’ to the last ‘few’ lines. However in exams these days all the stages (i.e. , to , !) are often ‘broken down’ for you and it might be best you work through the problem thoroughly and methodically.

• What indicator is used for this titration and what is the colour change at the end–point?

Starch indicator is used for the titration, when the last of the iodine reacts with the thiosulphate, the blue colour from the starch–iodine ‘complex’ is discharged and the solution becomes colourless.

25.  Calculate the molarities of iron(II) and iron(III) ions in a mixed solution from the following data.

• 0cm³ of the original mixture was acidified with dilute sulfuric acid and required 22.5 cm³ of 0.0200 mol dm^–3 KMnO₄ for complete oxidation.

 mol KMnO₄ = 0.0200 x 22.5 / 1000

               = 0.00045,

mol Fe²⁺ = mol KMnO₄ x 5

               = 0.00225 in 25 cm³,

scaling up to 1 dm³, molarity of the original Fe²⁺ 

                 = 0.00225 x 1000 / 25.0 

                = 0.090 mol dm^–3

• a further 0cm³ of the original iron(II)/iron(III) mixture was reduced with zinc and acid and it then required 37.6 cm³ of the KMnO₄ for complete oxidation.

The 2nd titration gives the total concentration of Fe²⁺ + Fe³⁺ because any Fe³⁺ has been reduced to Fe²⁺,

mol KMnO₄ = 0.0200 x 37.6 / 1000

         = 0.000752,

total mol Fe²⁺ titrated = mol KMnO₄ x 5

                     = 0.00376 in 25 cm³,

scaling up to 1 dm³, total molarity of  Fe²⁺ + Fe³⁺ in original solution = 0.00376 x 1000 / 25.0

               = 0.150 mol dm ^–3,

so using the result from (a) the  Fe³⁺ concentration 

     = ‘Fe’ total – Fe²⁺

           = 0.150 – 0.090 

       = 0.060 mol dm^–3

26. A piece of rusted iron was analysed to find out how much of the iron had been oxidised to rust [hydrated iron(III) oxide]. A small sample of the iron was dissolved in excess dilute sulfuric acid to give 250 cm³of solution. The solution contains Fe²⁺ions from the unrusted iron dissolving in the acid, and, Fe³⁺ ions from the rusted iron.

(a) 25.0 cm³ of this solution required 16.9 cm³ of 0.0200 mol dm^–3 KMnO₄ for complete oxidation of the Fe²⁺ ions.

Calculate the moles of Fe²⁺ ions in the sample titrated. 

mol Fe²⁺ = 5 x MnO₄^– 

= 5 x 0.0200 x 16.9 / 1000 

= 0.00169 mol 

= unreacted iron (which dissolved in the acid to form Fe²⁺).

• To a second 00cm³ of the rusted iron  solution an oxidising agent was added to convert all the Fe²⁺ ions present to Fe³⁺ ions. The Fe³⁺ ions were titrated with a solution of EDTA^4–_(aq) ions and 17.6 cm³ of 0.100 mol dm^–3 EDTA were required. 

mol Fe³⁺ = EDTA^4– = 0.100 x 17.6 / 1000 = 0.00176 mol = total mol iron in the sample titrated.

Assuming 1 mole of EDTA reacts with 1 mole of Fe³⁺ ions, calculate the moles of Fe³⁺ ions in the sample.

• From your calculations in (a) and (b) calculate the ratio of rusted iron to unrusted iron and hence the percentage of iron that had rusted.

calculation (a) gives the relative moles of unreacted iron Fe, as it dissolved to form the titratable Fe²⁺.

Calculation (b) gives the total of unreacted Fe + the rust i.e. Fe³⁺, because any Fe²⁺ formed from Fe has been oxidised to Fe³⁺.

So from the original mixture (in terms of the 25 cm³ sample), mol unreacted Fe = 0.00169,

mol of reacted iron = 0.00176 – 0.00169 = 0.00007.

Therefore the % rusted iron = 0.00007 x 100 / 0.00176 = 3.98 % rusted iron.

25. 0 cm³of an iodine solution was titrated with 0.100 mol dm^–3 sodium thiosulfate solution and the iodine reacted with 17.6 cm³ of the thiosulfate solution.

• give the reaction equation.

 I_2(aq) + 2S₂O₃^2–_(aq) ==> S₄O₆^2–_(aq) + 2I^–_(aq)           

or I_2(aq) + 2Na₂S₂O_3(aq) ==> Na₂S₄O_6(aq) + 2NaI_(aq)

• what indicator is used? and describe the end–point in the titration.

Starch indicator is used, starch gives a blue/black colour with iodine, this colour disappears when the last of the iodine is titrated, so a blue to colourless sharp end–point is observed.

• calculate the concentration of the iodine solution in mol dm^–3and g dm^–3.

mole ‘thio’ = 0.100 x 17.6/1000

                 = 0.00176,

              mol I₂ = 0.00176 ÷ 2

             = 0.00088 in 25 cm³,

scaling up gives 0.00088 x 1000 ÷ 25

              = 0.0352 mol dm^–3 for molarity of iodine,

            formula mass I₂ = 2 x 127

                                        = 254,

                            so concentration 

                                      = 0.0352 x 254  = 8.94 g dm^–

28. 01gof an impure sample of potassium dichromate (VI), K₂Cr₂O₇, was dissolved in dil. sulfuric acid and made up to 250 cm³ in a calibrated volumetric flask. A 25.0 cm³ aliquot of this solution pipetted into a conical flask and excess potassium iodide solution and starch indicator were added. The liberated iodine was titrated with 0.100 mol dm^–3 sodium thiosulfate and the starch turned colourless after 20.0 cm³ was added.

• Using the half–equations from Q3(a)(ii) and Q2(a)(ii), construct the full balanced equation for the reaction between the dichromate(VI) ion and the iodide ion.

Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6I^–_(aq) ==> 2Cr³⁺_(aq) + 3I_2(aq) + 7H₂O_(l)

• Using the half–equations from Q2(a) construct the balanced redox equation for the reaction between the thiosulfate ion and iodine.

2S₂O₃^2–_(aq)  +  I_2(aq)  ==>  S₄O₆^2–_(aq) + 2I^–_(aq)

• Calculate the moles of sodium thiosulfate used in the titration and hence the number of moles of iodine titrated.

               mol ‘thio’ = 20.0 x 0.100/1000

                                = 0.002,

therefore from equation 

                               mol iodine = mol ‘thio’/2

                                                  = 0.001

• Calculate the moles of dichromate(VI) ion that reacted to give the iodine titrated in the titration.

From equation (a) mol dichromate(VI) reacting

                                    = mol iodine liberated/3

                                       = 0.000333 (3sf)

• Calculate the formula mass of potassium dichromate(VI) and the mass of it in the 25.0 cm³aliquot titrated. 

M_r(K₂Cr₂O₇) = 294.2

mass K₂Cr₂O₇ titrated = 0.000333 x 294.2

                              =  0.0980 g (3 sf)

• Calculate the total mass of potassium dichromate(VI) in the original sample and hence its % purity.

Since the aliquot of 25.0 cm³ is 1/10th of the total solution in the flask, the total mass of the K₂Cr₂O₇ in original sample dissolved in the flask solution = 10 x 0.0980g

                             = 0.98g

and the % purity of the K₂Cr₂O₇ = 0.98 x 100/1.01

                            = 97.0 % (3 sf)

29. This question involves titrating ethanedioic acid (oxalic acid), H₂C₂O₄or (COOH)₂ (i) with standard sodium hydroxide solution and then with potassium manganate(VII) solution (potassium permanganate, KMnO₄).

The titration data is as follows:

10 cm³ of a H₂C₂O₄ solution required 8.50 cm³ of a 0.20 mol dm^-3 (0.20M) solution of sodium hydroxide for complete neutralisation using phenolphthalein indicator (first permanent pink end-point).

10 cm³ of the same H₂C₂O₄ solution required 8.20 cm³ of a KMnO₄ solution for complete oxidation to carbon dioxide and water in the presence of dilute sulfuric acid to further acidify the ethanedioic acid solution (first permanent pink end-point).

(a) Write an equation for the neutralisation reaction of ethanedioic acid with sodium hydroxide.

H₂C₂O₄  +  2NaOH  ===>  Na₂C₂O₄  +  2H₂O

or:  (COOH)₂  +  2NaOH  ===> (COONa)₂  +  2H₂O

 

• Calculate the moles of H₂C₂O₄in the solution and the molarity of the ethanedioic acid solution.

moles used in the NaOH in titration = 0.20 x 8.5/1000

                                                 = 0.0017 mol NaOH

from the equation mol H₂C₂O₄ = mol NaOH/2

therefore mol H₂C₂O₄ = 0.0017/2

                                    = 0.00085 in 10 cm³ of the acid solution.

volume of H₂C₂O₄ solution = 10/1000

                                             g= 0.01 dm³

therefore molarity of H₂C₂O₄ = 0.00085/0.01

                                      =  0.085 mol dm^-3 (0.085M)

(c) Given the following half-reactions:

(i)   MnO₄^–_(aq)  +  8H⁺_(aq)  +  5e^–  ===>  Mn²⁺_(aq)  +  4H₂O_(l) involves the gain of 5 electrons,

(ii)   H₂C₂O_4(aq)  – 2e^–   ===> 2CO_2(g)   +   2H⁺_(aq) 

involves the loss of 2 electrons, so to balance the electron gain and loss you need to add together 2 x (i) plus 5 x