A half-cell reaction represented by (i) as given below...

A half-cell reaction represented by (i) as given below $\mathrm{Fe}(\mathrm{OH})_2(\mathrm{~s})+2 e^{-} \rightarrow \mathrm{Fe}(\mathrm{~s})+2 \mathrm{OH}^{-}(\mathrm{aq}) \quad E^{\ominus}=-0.9 \mathrm{~V} \mathrm{(i)}$ takes place in two different electrochemical cells, I and II, in which the other half cell reactions are (ii) and (iii) respectively: $$\begin{array}{ll}A l^{3+}(a q)+3 e^{-} → A l(s) & E^{\ominus}=-1.7 \mathrm{~V} \mathrm{(ii)}\\ A g B r(s)+e^{-} → A g(s)+B r^{-}(a q) & E^{\ominus}=-0.07 \mathrm{~V} \mathrm{(iii)} \end{array}$$The correct option that represents the redox reactions in cells I and II is:$