JEE Main 2026 — Chemistry Physical Chemistry

For a spontaneous reaction, the cell potential $E^{\theta}_{cell}$Ecellθ​ 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:

$E^{\theta}_{AgBr/Ag} = +0.07$EAgBr/Agθ​=+0.07 V

$E^{\theta}_{Fe(OH)_2/Fe} = -0.88$EFe(OH)2​/Feθ​=−0.88 V

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

Cathode reaction: $2AgBr(s) + 2e^{-} \rightarrow 2Ag(s) + 2Br^{-}(aq)$2AgBr(s)+2e−→2Ag(s)+2Br−(aq)

Anode reaction: $Fe(s) + 2OH^{-}(aq) \rightarrow Fe(OH)_2(s) + 2e^{-}$Fe(s)+2OH−(aq)→Fe(OH)2​(s)+2e−

Overall cell reaction:

$$Fe(s) + 2OH^{-}(aq) + 2AgBr(s) \rightleftharpoons Fe(OH)_2(s) + 2Ag(s) + 2Br^{-}(aq)$$Fe(s)+2OH−(aq)+2AgBr(s)⇌Fe(OH)2​(s)+2Ag(s)+2Br−(aq)

Standard cell potential:

$E^{\theta}_{cell} = E^{\theta}_{cathode} - E^{\theta}_{anode} = 0.07 - (-0.88) = +0.95$Ecellθ​=Ecathodeθ​−Eanodeθ​=0.07−(−0.88)=+0.95 V

$E^{\theta}_{cell}$Ecellθ​ is an intensive property, and $Fe$Fe is oxidized in the cell. Thus, only the overall reaction given in the first option is correct.

Answer: Overall reaction $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightleftharpoons Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)$Fe(s)+2OH−(aq)+2AgBr(s)⇌Fe(OH)2​(s)+2Ag(s)+2Br−(aq)