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Question

For the cell reaction
2Fe3+(aq) + 2I(aq) 2Fe2+(aq) + I2(aq)E°cell=0.24 V at 298 K The standard Gibbs energy Go of the cell reaction is:
[Given that Faraday constant F = 96500 C mol–1]

A
– 46.32 kJ mol1
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B
– 23.16 kJ mol1
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C
46.32 kJ mol1
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D
23.16 kJ mol1
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Solution

The correct option is A – 46.32 kJ mol1
Go = -nF Eocell
= -2 ×96500×0.24 J mol1
= -46320 J mol1
= -46.32 kJ mol1

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