JEE Main 2026 — Electrochemistry Question with Solution

For a spontaneous reaction, the cell potential $E_{cell}^{\theta }$ 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:
$E_{AgBr/Ag}^{\theta }=+0.07$ V
$E_{Fe(OH{)}_2/Fe}^{\theta }=-0.88$ V

Since $+0.07$ V $>-0.88$ V, $AgBr$ undergoes reduction at the cathode and $Fe$ undergoes oxidation at the anode.

Cathode reaction: $2AgBr(s)+2e^{-}\to 2Ag(s)+2B{r}^{-}(aq)$
Anode reaction: $Fe(s)+2O{H}^{-}(aq)\to Fe(OH{)}_2(s)+2e^{-}$

Overall cell reaction:
$Fe(s)+2O{H}^{-}(aq)+2AgBr(s)\rightleftharpoons Fe(OH{)}_2(s)+2Ag(s)+2B{r}^{-}(aq)$

Standard cell potential:
$E_{cell}^{\theta }=E_{cathode}^{\theta }-E_{anode}^{\theta }=0.07-(-0.88)=+0.95$ V

$E_{cell}^{\theta }$ 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 $Fe(s)+2O{H}^{-}(aq)+2AgBr(s)\rightleftharpoons Fe(OH{)}_2(s)+2Ag(s)+2B{r}^{-}(aq)$