For the cell, ${ }^{\ominus} \mathrm{Zn}_{(s)}\left|\mathrm{Zn}^{+2}(1 \mathrm{M})\right|\left|\mathrm{Ag}^{+1}(1 \mathrm{M})\right| \mathrm{Ag}_{(s)}^{\oplus}$ If concentration of $\mathrm{Zn}^{+2}$ decreases to 0.1 M at 298 K . then emf of cell

The electrochemical cell is represented as $\mathrm{Zn_{(s)}} | \mathrm{Zn^{2+}} (1\ \mathrm{M}) \parallel \mathrm{Ag^+} (1\ \mathrm{M}) | \mathrm{Ag_{(s)}}$.

Cell Reactions and Electron Transfer
At the anode: $\mathrm{Zn_{(s)}} \rightarrow \mathrm{Zn^{2+}_{(aq)}} + 2e^{-}$
At the cathode: $2\mathrm{Ag^+_{(aq)}} + 2e^{-} \rightarrow 2\mathrm{Ag_{(s)}}$
The overall cell reaction is $\mathrm{Zn_{(s)}} + 2\mathrm{Ag^+_{(aq)}} \rightarrow \mathrm{Zn^{2+}_{(aq)}} + 2\mathrm{Ag_{(s)}}$, with $n = 2$ electrons transferred.

Applying the Nernst Equation
The Nernst equation at 298 K is $E_\text{cell} = E^0_\text{cell} - \frac{0.0592}{n} \log Q$, where the reaction quotient $Q = \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Ag^+}]^2}$.

Initially, with $[\mathrm{Zn^{2+}}] = 1\ \mathrm{M}$ and $[\mathrm{Ag^+}] = 1\ \mathrm{M}$, $Q_1 = 1$ and $\log Q_1 = 0$, giving $E_\text{cell,1} = E^0_\text{cell}$.

After the change, $[\mathrm{Zn^{2+}}] = 0.1\ \mathrm{M}$ and $[\mathrm{Ag^+}] = 1\ \mathrm{M}$, so $Q_2 = 0.1$ and $\log Q_2 = -1$. Substituting gives $E_\text{cell,2} = E^0_\text{cell} - \frac{0.0592}{2} (-1) = E^0_\text{cell} + 0.0296\ \mathrm{V}$.

The change in EMF is $\Delta E_\text{cell} = E_\text{cell,2} - E_\text{cell,1} = (E^0_\text{cell} + 0.0296) - E^0_\text{cell} = +0.0296\ \mathrm{V}$, indicating an increase.

The answer is $\boxed{\text{C}}$.