Consider the Daniell cell, $\mathrm{Zn}\left|\mathrm{Zn} \mathrm{SO}_{4(\mathrm{IM})}\right|\left|\m

Question

Consider the Daniell cell, $\mathrm{Zn}\left|\mathrm{Zn} \mathrm{SO}_{4(\mathrm{IM})}ight|\left|\mathrm{Cu} \mathrm{SO}_{4(\mathrm{IM})}ight| \mathrm{Cu}$ and choose the electrode reactions from the following\ i) $\mathrm{Zn}_{(\mathrm{IM})}^{2+}+2 \mathrm{e}^{-} ightarrow \mathrm{Zn}_{(\mathrm{s})}$ ii) $\mathrm{Zn}_{(\mathrm{s})} ightarrow \mathrm{Zn}_{(\mathrm{IM})}^{2+}+2 \mathrm{e}^{-}$\ iii) $\mathrm{Cu}_{(\mathrm{IM})}^{2+}+2 \mathrm{e}^{-} ightarrow \mathrm{Cu}_{(\mathrm{s})}$\ iv) $\mathrm{Cu}_{(\mathrm{s})} ightarrow \mathrm{Cu}_{(\mathrm{IM})}^{2+}+2 \mathrm{e}^{-}$ The reaction at the anode and cathode are respectively\

Professor PC Thomas Classes · Answer

Oxidation at the L.H.S. electrode, (anode), $\mathrm{Zn} ightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$\ Reduction at the R.H.S. electrode (cathode) $\mathrm{Cu}^{2+} 2 \mathrm{e}^{-} ightarrow \mathrm{Cu}$\