Question
An electrochemical cell is constructed using half cells in the direction of spontaneous change Fe(OH)_2(s) + 2e^- Fe(s) + 2OH^-(aq) E^ = -0.88 V and AgBr(s) + e^- Ag(s) + Br^-(aq) E^ = +0.07 V Which of the following option is correct ?
JEE Main
Chemistry
Electrochemistry
2026
JEE Main 2026 (02 April Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. Overall reaction Fe(s) + 2OH^-(aq) + 2AgBr(s) Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)
- B. E^ _ cell = -0.95 V
- C. Fe is reduced in the electrochemical cell
- D. E^ _ cell is an extensive property
Answer
A. Overall reaction Fe(s) + 2OH^-(aq) + 2AgBr(s) Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)
Step-by-step solution
For a spontaneous reaction, the cell potential E^ _ cell must be positive. The half-cell with the higher standard reduction potential acts as the cathode (reduction), and the one with the lower standard reduction potential acts as the anode (oxidation). Given standard reduction potentials: E^ _ AgBr/Ag = +0.07 V E^ _ Fe(OH)_2/Fe = -0.88 V Since +0.07 V > -0.88 V, AgBr undergoes reduction at the cathode and Fe undergoes oxidation at the anode. Cathode reaction: 2AgBr(s) + 2e^ - 2Ag(s) + 2Br^ - (aq) Anode reaction: Fe(s) + 2OH^ - (aq) Fe(OH)_2(s) + 2e^ - Overall cell reaction: Fe(s) + 2OH^ - (aq) + 2AgBr(s) Fe(OH)_2(s) + 2Ag(s) + 2Br^ - (aq) Standard cell potential: E^ _ cell = E^ _ cathode - E^ _ anode = 0.07 - (-0.88) = +0.95 V E^ _ cell is an intensive property, and Fe is oxidized in the cell. Thus, only the overall reaction given in the first option is correct. Answer: Overall reaction Fe(s) + 2OH^-(aq) + 2AgBr(s) Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)
Related: Chemistry — Electrochemistry · All PYQ Banks