Calculate Magnitude of Cell Potential for Changed Concentration

For a concentration cell represented as:
$\text{M(s)} \mid \text{M}^+(\text{aq; } c_1) \mid\mid \text{M}^+(\text{aq; } c_2) \mid \text{M(s)}$

The cell reaction is:
$\text{M}^+(\text{aq; } c_2) \rightleftharpoons \text{M}^+(\text{aq; } c_1)$

Since both half-cells consist of the same metal-metal ion electrode, the standard cell potential $E^\circ_{\text{cell}} = 0\text{ V}$.

Using the Nernst equation for the cell potential at temperature $T$:
$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} \left( \frac{c_1}{c_2} \right)$

For $n = 1$:
$|E_{\text{cell}}| = \frac{2.303 RT}{F} \log_{10} \left( \frac{c_2}{c_1} \right)$

Case 1:
Initial concentration $c_1 = 0.05\text{ M}$ and $c_2 = 1\text{ M}$.
$|E_{\text{cell}, 1}| = \frac{2.303 RT}{F} \log_{10} \left( \frac{1}{0.05} \right) = \frac{2.303 RT}{F} \log_{10}(20) = 70\text{ mV}$

Case 2:
New concentration $c_1' = 0.0025\text{ M}$ and $c_2 = 1\text{ M}$.
$|E_{\text{cell}, 2}| = \frac{2.303 RT}{F} \log_{10} \left( \frac{1}{0.0025} \right) = \frac{2.303 RT}{F} \log_{10}(400)$

Since $400 = 20^2$, we can simplify the logarithmic term:
$\log_{10}(400) = \log_{10}(20^2) = 2 \log_{10}(20)$

Substituting this back into the equation:
$|E_{\text{cell}, 2}| = 2 \times \left( \frac{2.303 RT}{F} \log_{10}(20) \right) = 2 \times |E_{\text{cell}, 1}|$

Substituting $|E_{\text{cell}, 1}| = 70\text{ mV}$:
$|E_{\text{cell}, 2}| = 2 \times 70\text{ mV} = 140\text{ mV}$

Thus, the correct option is C.