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Q.

Given electrode potential are
Fe3+ + e → Fe2+; E° = 0.771 V, I2 + 2e → 2I;
E° = 0.536 V, E°cell for the cell reaction
2Fe3+ + 2I → 2Fe2+ + I2 is

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a

0.771 – 0.5 × 0.536 = 0.503 V

b

0.536 – 0.771 = – 0.235 V

c

0.771 – 0.536 = 0.235 V

d

2 × 0.771 – 0.536 = 1.006 V

answer is C.

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Detailed Solution

Higher reduction potential acts as cathode and lower reduction potential acts as anode

Ecell°=Ec°-Ea°= 0.771 - 0.536 =0.235V

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