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(A) reaction is spontaneous if $E^{0}_{cell} > 0$. Given half-reactions are: $Fe(OH)_{2}(s) + 2e^{-} \rightarrow Fe(s) + 2OH^{-}(aq)$ $(E^{0}_{red} =

Vedclass · Accepted Answer

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

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(A) reaction is spontaneous if $E^{0}_{cell} > 0$. Given half-reactions are: $Fe(OH)_{2}(s) + 2e^{-} ightarrow Fe(s) + 2OH^{-}(aq)$ $(E^{0}_{red} = -0.88 	ext{ V})$ $AgBr(s) + e^{-} ightarrow Ag(s) + Br^{-}(aq)$ $(E^{0}_{red} = +0.07 	ext{ V})$ For a spontaneous reaction, the half-reaction with the higher $E^{0}_{red}$ acts as the cathode (reduction) and the one with the lower $E^{0}_{red}$ acts as the anode (oxidation). Anode (Oxidation): $Fe(s) + 2OH^{-}(aq) ightarrow Fe(OH)_{2}(s) + 2e^{-}$ $(E^{0}_{ox} = +0.88 	ext{ V})$ Cathode (Reduction): $2AgBr(s) + 2e^{-} ightarrow 2Ag(s) + 2Br^{-}(aq)$ $(E^{0}_{red} = +0.07 	ext{ V})$ Overall reaction: $Fe(s) + 2OH^{-}(aq) + 2AgBr(s) ightarrow Fe(OH)_{2}(s) + 2Ag(s) + 2Br^{-}(aq)$ $E^{0}_{cell} = E^{0}_{cathode} + E^{0}_{ox} = 0.07 	ext{ V} + 0.88 	ext{ V} = 0.95 	ext{ V}$. Since $E^{0}_{cell} > 0$, the reaction is spontaneous. Thus, option $A$ is correct.

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