$EMF = {E_{cathode}} - {E_{anode}}$$ = 0.75 - (0.5)$; $EMF = 0.25\,V$.
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| Column $I$ | Column $II$ |
| $(A)$ $CH_3COONa$ | $(i)$ Almost neutral, $pH > 7$ or $< 7$ |
| $(B)$ $NH_4Cl$ | $(ii)$ Acidic, $pH < 7$ |
| $(C)$ $NaNO_3$ | $(iii)$ Alkaline, $pH > 7$ |
| $(D)$ $CH_3COONH_4$ | $(iv)$ Neutral, $pH = 7$ |
Product will be
$(I)$ $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}$
$(II)$ $ \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
$(III)$ $\mathrm{Na}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\left(\Delta_{0}>\mathrm{P}\right)$
$(IV)$ $\left(\mathrm{Et}_{4} \mathrm{N}\right)_{2}\left[\mathrm{CoCl}_{4}\right]$