Question

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

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

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): $C{u}^{2+}+2e^{-}\to Cu$
Anode (oxidation): $Sn\to S{n}^{2+}+2e^{-}$
In the representation of the cell, anode keeps on the left and cathode on the right.
Hence, the correct representation of the cells
$Sn(s)\left|S{n}^{2+}(1.0M)\right|\left|C{u}^{2+}(1.0M)\right|Cu(s)$