To understand how the concentration of ions affects the cell potential, we will first use the Nernst equation. The Nernst equation gives us a way to calculate the potential of a cell under non-standard conditions and is represented as follows:

$$ E = E^0 - \frac{RT}{nF} \ln \frac{a_{\text{Red}}}{a_{\text{Ox}}} $$

In our case, we're analyzing a concentration cell where the metal M is the same in both the anode and the cathode but with different concentrations of M⁺. Here, the standard potential $ E^0 $ is zero because the same substance is used as both the anode and cathode.

The cell reaction becomes:

$$ \text{M}(s) | \text{M}^{+}(0.05 \text{ M at anode}) \Longrightarrow \text{M}(s) | \text{M}^{+}(1 \text{ M at cathode}) $$

Since $ E^0 = 0 $, the Nernst equation simplifies to:

$$ E = - \frac{RT}{nF} \ln \frac{[\text{M}^{+} \text{ (cathode)}]}{[\text{M}^{+} \text{ (anode)}]} $$

Which turns into:

$$ E = - \frac{RT}{nF} \ln \frac{1}{0.05} , $$

and we can further simplify using $ \ln(\frac{1}{0.05}) = -\ln(0.05) $. Assume the reaction involves the transfer of 1 mole of electrons (n=1), the value of F (Faraday constant) is approximately 96485 C/mol, and the temperature T is 298K (room temperature). The gas constant R is 8.314 J/(mol·K). Plugging in these values:

$$ E \approx - \frac{(8.314 \, \text{J/mol·K})(298 \, \text{K})}{(1)(96485 \, \text{C/mol})} \ln(0.05) $$

This equation can be used to find the potential in volts when the concentration at anode was 0.05 M and was resulting in 70 mV.

Now, switching the anode concentration to 0.0025 M, we reapply the Nernst equation:

$$ E = - \frac{RT}{nF} \ln \frac{1}{0.0025} $$

We can see the ratio of concentrations has changed, moving from 1/0.05 to 1/0.0025. Thus, the concentration difference across the membrane has increased, which should increase the voltage according to the Nernst equation. Specifically, the ratio changed by a factor of 0.05/0.0025 = 20 times.

Given that $\ln(\frac{1}{x})$ is proportional to the voltage, when the ratio of concentration changes by twenty-fold, and because $\ln(0.0025) = -\ln(400)$ which is twice the $\ln(20)$, the potential will double. Since the initial 70 mV doubles, the new cell potential will be:

$$ E \approx 2 \times 70 \, \text{mV} = 140 \, \text{mV}. $$

Therefore, the correct answer is:

Option C: 140 mV