Question

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Answer 1

Answer:

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 2

Answer:

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