Question
Given at 298 K: E^ _ Fe^ 2+ /Fe = X Volt; E^ _ Fe^ 3+ /Fe = Y Volt. The E^ _ Fe^ 3+ /Fe^ 2+ in Volt at 298 K is given by:
JEE Main
Chemistry
Electrochemistry
2026
JEE Main 2026 (08 April Shift 2)
JEE Main Chemistry Question (2026) — Solution
Options
- A. 2X-3Y
- B. 3Y-2X
- C. 3Y+2X
- D. Y+X
Answer
B. 3Y-2X
Step-by-step solution
The half-reactions and their standard Gibbs free energy changes are given by: Fe^ 2+ + 2e^ - Fe G^ _ 1 = -2FX Fe^ 3+ + 3e^ - Fe G^ _ 2 = -3FY The required half-reaction is: Fe^ 3+ + e^ - Fe^ 2+ G^ _ 3 = -1FE^ _ Fe^ 3+ /Fe^ 2+ This reaction is obtained by subtracting the first reaction from the second reaction: G^ _ 3 = G^ _ 2 - G^ _ 1 -FE^ _ Fe^ 3+ /Fe^ 2+ = -3FY - (-2FX) -FE^ _ Fe^ 3+ /Fe^ 2+ = -3FY + 2FX E^ _ Fe^ 3+ /Fe^ 2+ = 3Y - 2X Answer: 3Y-2X
Related: Chemistry — Electrochemistry · All PYQ Banks