Which from following statement is correct regarding emf of cell for the cell reaction, $\mathrm{Cd}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{+2} \longrightarrow \mathrm{Cd}_{(\mathrm{aq})}^{+2}+\mathrm{Cu}_{(\mathrm{s})}$ if concentration of $\mathrm{Cd}^{+2}$ is 10 times greater than concentration of $\mathrm{Cu}_{(\mathrm{aq})}^{+2}$ at 298 K ?

The cell reaction is:
$\mathrm{Cd}_{(\mathrm{s})} + \mathrm{Cu}_{(\mathrm{aq})}^{+2} \longrightarrow \mathrm{Cd}_{(\mathrm{aq})}^{+2} + \mathrm{Cu}_{(\mathrm{s})}$

Applying the Nernst equation at 298 K:
$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0592}{n} \log Q$

Two electrons are transferred, so $n = 2$. The reaction quotient is $Q = \frac{[\mathrm{Cd}^{+2}]}{[\mathrm{Cu}^{+2}]}$.

Given that $[\mathrm{Cd}^{+2}] = 10 \times [\mathrm{Cu}^{+2}]$, substitution yields $Q = 10$.

Substituting $n=2$ and $Q=10$:

$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0592}{2} \log(10)$

Since $\log(10) = 1$:

$E_{\text{cell}} = E_{\text{cell}}^{\circ} - 0.0296 \text{ V}$

Thus, $E_{\text{cell}}$ is less than $E_{\text{cell}}^{\circ}$ by 0.0296 V. The correct option is D.