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

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.

____________________________________________________

<Judge It Yourself...>

Hope it helps you! ヅ

✪ Be Brainly ✪

Answer 2

ANSWER

At anode :          [Sn2+(aq)           →        Sn4+(aq)+2e−]×5 

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

Cell reaction : 

_____________________________________________________________________

2MnO4−(aq)+5Sn2+(aq)+16H+(aq)→2Mn2+(aq)+5Sn4+(aq)+8H2O(l) 

______________________________________________________________________

Ecell∘=Ecathode∘−Eanode∘=1.51V−0.15V 

                                              =1.36V 

As cell potential is positive therefore the reaction is product favoured.

HOPE IT'S HELPFUL FOR YOU.