Question

Two half cell reactions of an electrochemical cell are given below :
MnO–4(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = + 1.51 V
Sn2+ (aq) → 4 Sn4+ (aq) + 2e–, E° = + 0.15 V
Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation...​

Answer 1

Explanation:

Two Half Cell Reactions in an Electrochemical cell :

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O E°cell = +1.51V

Sn2+ → Sn2+ + 2e- E°cell = 0.15V

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Multiply first reaction by 2 and second reaction by 5 to cancel out the electrons, and adding, the redox reaction will be,

2MnO4- + 5Sn2+ + 16H+ → 5Sn4+ + 2Mn2+ + 8H2O

Oxidation potential of 2nd reaction = -0.15 V (as given value is the reduction potential).

Hence, E°cell = 1.51 + (-0.15) V = 1.36 V

As overall cell potential is positive, the reaction will favour formation of products as represented in overall cell reaction.

Answer 2

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The reactions can be represented at anode and at cathode in the following ways :

At anode (oxidation) :

Sn²⁺ → = Sn⁴⁺ (aq) + 2e⁻ ] × 5 E° = + 0.15 V

At cathode (reduction) :

MnO₄⁻(aq) + 8H⁺ (aq) + 5e⁻ → Mn²⁺ (aq) + 4H₂O (I)] × 2 E° = + 1.51 V

The Net R × M = 2MnO₄⁻(aq) + 16H⁺ + 5Sn²⁺ →

2Mn²⁺ + 5Sn⁴⁺ + 8H₂O

Now E°cell = E°cathode – E°anode

= 1.51 – 0.15 = + 1.36 V

∴ Positive value of E°cell favours formation of product.

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