Correct Option (A)
Explanation :
The cell is represented as
$\mathrm{Tl}_{(s)}\left|\mathrm{Tl}^{+}(1.0 \mathrm{M}) \| \mathrm{Sn}^{4+}(1.0 \mathrm{M}), \mathrm{Sn}^{2+}(1.0 \mathrm{M})\right| \mathrm{Pt}$
$\text { The cell reaction is } \quad\left(\mathrm{Tl}_{(s)} \longrightarrow \mathrm{Tl}^{+}+e^{-}\right) \times 2$
$\mathrm{Sn}^{4+}+2 e^{-} \longrightarrow \mathrm{Sn}^{2+}$
$\text { Overall reaction }: 2 \mathrm{Tl}_{(s)}+\mathrm{Sn}^{4+} \longrightarrow 2 \mathrm{Tl}^{+}+\mathrm{Sn}^{2+}$
$E=\left(E_{\text {Right }}^{\mathrm{o}}-E_{\text {Left }}^{\mathrm{o}}\right)-\frac{0.0592}{2} \log \frac{\left[\mathrm{Tl}^{+}\right]^2\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Sn}^{4+}\right]}$
$=0.47 \mathrm{~V}-0.0296 \log (10)^2$
${[\because \text { Tl concentration increases tenfold] }}$
$=0.47-0.0592=0.411 \mathrm{~V}$