For the cell reaction 2Fe3+(aq)+2I−(aq)→2Fe2+(aq)+I2(aq)E°cell=0.24Vat298K The standard Gibbs energy △Go of the cell reaction is:
[Given that Faraday constant F = 96500 C mol–1]
A
– 46.32 kJ mol−1
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B
– 23.16 kJ mol−1
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C
46.32 kJmol−1
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D
23.16 kJmol−1
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Solution
The correct option is A – 46.32 kJ mol−1 △Go = -nFEocell
= -2 ×96500×0.24 J mol−1
= -46320 J mol−1
= -46.32 kJmol−1