For the cell reaction 2Fe3+(aq) + 2I-(aq) 2Fe2+(aq) + I2(aq) E cell = 0.24 V at 298 K. The standard Gibbs energy ( r G ) of the cell reaction is : [Given that Faraday constant F = 96500 C mol-1]
- 46.32 kJ mol-1
- 23.16 kJ mol-1
- -46.32 kJ mol-1
- -23.16 kJ mol-1
Answer
(3) -46.32 kJ mol-1
Solution
Half reactions: 2Fe3+ + 2e- 2Fe2+ and 2I- I2 + 2e-
n = 2
As per NCERT, r G = -nF E cell
= -2 96500 0.24 J mol-1
= -46320 J mol-1
= -46.32 kJ mol-1
Answer: (3)