Step 1: Understanding the Concept:
The cell potential (EMF) changes with concentration as described by the Nernst Equation.
Step 2: Key Formula or Approach:
$\text{E}_{\text{cell}} = \text{E}^\circ_{\text{cell}} - \frac{0.0592}{\text{n}} \log\left(\frac{[\text{Products}]}{[\text{Reactants}]}\right)$
Cell reaction: $\text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag}$, where $\text{n}=2$.
Step 3: Detailed Explanation:
Initial state: $[\text{Zn}^{2+}] = 1\text{ M}$, $[\text{Ag}^+] = 1\text{ M} \rightarrow \text{E}_1 = \text{E}^\circ_{\text{cell}}$.
New state: $[\text{Zn}^{2+}] = 0.1\text{ M}$, $[\text{Ag}^+] = 1\text{ M}$.
\[ \text{E}_2 = \text{E}^\circ_{\text{cell}} - \frac{0.0592}{2} \log\left(\frac{0.1}{1^2}\right) \]
\[ \text{E}_2 = \text{E}^\circ_{\text{cell}} - 0.0296 \times \log(10^{-1}) \]
\[ \text{E}_2 = \text{E}^\circ_{\text{cell}} - 0.0296 \times (-1) = \text{E}^\circ_{\text{cell}} + 0.0296\text{ V} \]
$\Delta \text{E} = \text{E}_2 - \text{E}_1 = +0.0296\text{ V}$.
Step 4: Final Answer:
The EMF increases by $0.0296\text{ V}$.