An electrochemical cell is constructed using half cells in the...

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

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

Cathode reaction: $2AgBr(s) + 2e^{-} \rightarrow 2Ag(s) + 2Br^{-}(aq)$
Anode reaction: $Fe(s) + 2OH^{-}(aq) \rightarrow Fe(OH)_2(s) + 2e^{-}$

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

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

$E^{\theta}_{cell}$ 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 $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightleftharpoons Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)$