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...
Answers
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) :
MnO–4(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO–4(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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ANSWER
At anode : [Sn2+(aq) → Sn4+(aq)+2e−]×5
At cathode : [MnO4−(aq)+8H+(aq)+5e− → Mn2+(aq)+4H2O(l)]×2
Cell reaction :
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2MnO4−(aq)+5Sn2+(aq)+16H+(aq)→2Mn2+(aq)+5Sn4+(aq)+8H2O(l)
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Ecell∘=Ecathode∘−Eanode∘=1.51V−0.15V
=1.36V
As cell potential is positive therefore the reaction is product favoured.
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