Question

A cell is constructed by coupling the two electrodes $Sn / Sn ^{2+}$ and $C u / C u^{2+}$. If $E^{o}\left(S n^{2+}, S n\right), E^{o}\left(C u^{2+}, C u\right)$ and $E^{o}$ (cell) are $-0.14 V, 0.34 V$ and $0.48 V$ respectively, the correct representation of the cell is

• A. $ Sn_{(s)}|S{{n}^{2+}}(0.1\,M)||C{{u}^{2+}}(1.0\,M)|Cu_{(s)} $
• B. $ Sn_{(s)}|S{{n}^{2+}}(1.0\,M)||C{{u}^{+}}(1.0\,M)|Cu_{(s)} $
• C. $ Sn_{(s)}|S{{n}^{2+}}(1.0\,M)||C{{u}^{2+}}(1.0\,M)|Cu_{(s)} $
• D. $ Cu_{(s)}|C{{u}^{2+}}(1.0\,M)||S{{n}^{2+}}(1.0\,M)|Sn_{(s)} $

Answer 1

Answer: (C)

Lower value of standard reduction potential represents anode while higher value of standard reduction potential represents cathode.
Hence, copper $(Cu)$ electrode acts as cathode and tin $(Sn)$ electrode acts as anode. Half-cell reaction:
Cathode (reduction): $C u^{2+}+2 e^{-} \rightarrow C u$
Anode (oxidation): $S n \rightarrow S n^{2+}+2 e^{-}$
In the representation of the cell, anode keeps on the left and cathode on the right.
Hence, the correct representation of the cells
$S n(s)\left|S n^{2+}(1.0 M)\right|\left|C u^{2+}(1.0 M)\right| C u(s)$