Question

A galvanic cell is constructed as follows. A half-cell consists of a platinum wire immersed in a solution containing $1.0 {M}$ of ${Sn}^{2+}$ and $1.0 {M}$ of ${Sn}^{4+}$, and another half-cell has a thallium rod immersed in a $1.0 {M}$ solution of ${Tl}^{+}$.

Given —

$: \mathrm{Sn}_{(a q)}^{4+}+2 e^{-} \longrightarrow \mathrm{Sn}_{(a q)}^{2+} ; \quad E^{\circ}=+0.13 \mathrm{~V}$ and $\mathrm{Tl}_{(a q)}^{+}+e^{-} \longrightarrow \mathrm{Tl}_{(s)} ; \quad E^{\circ}=-0.34 \mathrm{~V}$,

What is the cell voltage if the $\mathrm{Tl}^{+}$concentration is increased tenfold?

(a) $0.411 \mathrm{~V}$

(b) $4.101 \mathrm{~V}$

(c) $0.492 \mathrm{~V}$

(d) $0.222 \mathrm{~V}$

Answer 1

Correct Option (A)

Explanation :

The cell is represented as

$\mathrm{Tl}_{(s)}\left|\mathrm{Tl}^{+}(1.0 \mathrm{M}) \| \mathrm{Sn}^{4+}(1.0 \mathrm{M}), \mathrm{Sn}^{2+}(1.0 \mathrm{M})\right| \mathrm{Pt}$

$\text { The cell reaction is } \quad\left(\mathrm{Tl}_{(s)} \longrightarrow \mathrm{Tl}^{+}+e^{-}\right) \times 2$

$\mathrm{Sn}^{4+}+2 e^{-} \longrightarrow \mathrm{Sn}^{2+}$

$\text { Overall reaction }: 2 \mathrm{Tl}_{(s)}+\mathrm{Sn}^{4+} \longrightarrow 2 \mathrm{Tl}^{+}+\mathrm{Sn}^{2+}$

$E=\left(E_{\text {Right }}^{\mathrm{o}}-E_{\text {Left }}^{\mathrm{o}}\right)-\frac{0.0592}{2} \log \frac{\left[\mathrm{Tl}^{+}\right]^2\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Sn}^{4+}\right]}$

$=0.47 \mathrm{~V}-0.0296 \log (10)^2$

${[\because \text { Tl concentration increases tenfold] }}$

$=0.47-0.0592=0.411 \mathrm{~V}$