Question

Consider the cell,

$Zn|Z{n}^{2+}\left(aq\right)\left(1.0M\right)||C{u}^{2+}\left(aq\right)\left(1.0M\right)|Cu$

The standard reduction potentials are $+0.35V$ for
$2e^{-}+C{u}^{2+}\left(aq\right)\to C{u}^{2+}\left(aq\right)\left(1.0M\right)|Cu$

The standard reduction potentials are $+0.35V$ for
$2e^{-}+C{u}^{2+}\left(aq\right)\to Cu$ and $-0.763V$ for $2e^{-}+Z{n}^{2+}\left(aq\right)\to Zn$
Calculate the emf of the cell and tell whether the cell reaction spontaneous or not ?

• A. 2.13 V, Non-Spontaneous
• B. 1.13 V, Spontaneous
• C. 2.13 V, Spontaneous
• D. 1.13 V, Non-Spontaneous

Answer 1

The correct option is B 1.13 V, Spontaneous

$E_{C{u}^{2+},Cu}^o=+0.35V$
$E_{Z{n}^{2+},Zn}^o=-0.763V$
Cell reactions are as follows :
RHS electrode: $C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$ (reduction)
LHS electrode: $Zn\to Z{n}^{2+}\left(aq\right)+2e^{-}$
(oxidation)

(ii) $E_{cell}=E_{\text{right electrode}}-E_{\text{left electrode}}$

$E_{cell}=E_{C{u}^{2+},Cu}^o+\frac{0.0591}{2}log\left[C{u}^{2+}\right]-E_{Z{n}^{2+},Zn}^o\frac{0.0591}{2}log\left[Z{n}^{2+}\right]$

$=+0.35+\frac{0.0591}{2}log\left(1\right)-(-0.763)+\frac{0.0591}{2}log\left(1\right)$

$=0.35+0.763$
$=1.13V$
Since $E_{cell}$ is positive the cell reactions as mentioned above are spontaneous. Thus option (b) is correct.