Given at 298 K : E Fe2+/Fe = X Volt E Fe3+/Fe = Y Volt The E Fe3+/Fe2+ in Volt at 298 K is given by :
- 2X − 3Y
- 3Y − 2X
- 3Y + 2X
- Y + X
Answer
(2) 3Y − 2X
Solution
Electrode potentials are not additive, but G = -nFE values are.
(i) Fe3+ + 3e- Fe; G1 = -3FY
(ii) Fe2+ + 2e- Fe; G2 = -2FX
(i) − (ii): Fe3+ + e- Fe2+; G3 = -3FY + 2FX
⇒ -1 F E Fe3+/Fe2+ = -3FY + 2FX
∴ E Fe3+/Fe2+ = 3Y - 2X
Answer: (2)