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Question
For the cell reaction
\[\ce{2Fe^{3+}_{( aq)} + 2l^-_{( aq)} -> 2Fe^{2+}_{( aq)} + l2_{( aq)}}\]
\[\ce{E^0_{cell}}\] = at 298 K. The standard Gibbs energy (∆G°) of the cell reactions is:
Options
– 46.32 KJ mol−1
– 23.16 KJ mol−1
46.32 KJ mol−1
23.16 KJ mol−1
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Solution
– 46.32 KJ mol−1
Explanation:
n = 2; \[\ce{E^0_{cell}}\] = 0.24 V; ∆G° = ?; F = 96500 C
∆G° = – nFE°
∆G°= – 2 × 96500 × 0.24
∆G° = – 46320 J mol−1
∆G° = – 46.32 KJ mol−1
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