Q.57. An electrochemical cell is constructed using half cells in the direction of spontaneous change Fe(OH)(s) + 2e Fe(s) + 2OH(aq) E = -0.88V and AgBr(s) + e Ag(s) + Br(aq) E = +0.07V Which of the

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Q.57. An electrochemical cell is constructed using half cells in the direction of spontaneous change Fe(OH)(s) + 2e  Fe(s) + 2OH(aq) E = -0.88V and AgBr(s) + e  Ag(s) + Br(aq) E = +0.07V Which of the following option is correct ?

FirstInTest · Accepted Answer

Correct answer: A. Given half-cell reactions and their standard electrode potentials:
1.  Fe(OH)$_2$(s) + 2e$^-$ $ightarrow$ Fe(s) + 2OH$^-$(aq); E$^{\circ}_1$ = -0.88V
2.  AgBr(s) + e$^-$ $ightarrow$ Ag(s) + Br$^-$(aq); E$^{\circ}_2$ = +0.07V

For a spontaneous reaction, the cell potential E$^{\circ}_{cell}$ must be positive. E$^{\circ}_{cell}$ = E$^{\circ}_{cathode}$ - E$^{\circ}_{anode}$. The species with the higher (more positive) reduction potential will act as the cathode, and the species with the lower (more negative) reduction potential will act as the anode.
Here, E$^{\circ}_2$ (+0.07V) > E$^{\circ}_1$ (-0.88V). So, AgBr/Ag will be the cathode, and Fe(OH)$_2$/Fe will be the anode.

At Cathode (reduction): $2 	imes (\mathrm{AgBr(s)} + \mathrm{e^-} ightarrow \mathrm{Ag(s)} + \mathrm{Br^-(aq)})$ (Multiply by 2 to balance electrons)
At Anode (oxidation): $\mathrm{Fe(s)} + 2\mathrm{OH^-(aq)} ightarrow \mathrm{Fe(OH)_2(s)} + 2\mathrm{e^-}$ (Reverse reaction 1)

Overall reaction: $\mathrm{Fe(s)} + 2\mathrm{OH^-(aq)} + 2\mathrm{AgBr(s)} ightarrow \mathrm{Fe(OH)_2(s)} + 2\mathrm{Ag(s)} + 2\mathrm{Br^-(aq)}$. This matches option (1).

Let's check other options:
(2) E$^{\circ}_{cell}$ = E$^{\circ}_{cathode}$ - E$^{\circ}_{anode}$ = E$^{\circ}_2$ - E$^{\circ}_1$ = (+0.07V) - (-0.88V) = +0.95V. Option (2) states E$^{\circ}_{cell}$ = -0.95V, which is incorrect.
(3) Fe is oxidized at the anode (Fe $ightarrow$ Fe(OH)$_2$), not reduced. So, option (3) is incorrect.
(4) E$^{\circ}_{cell}$ is an intensive property, not an extensive property. So, option (4) is incorrect.

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