Step-by-Step Solution:

1. Identify the type of cell and relevant equation. The electrochemical cell described is a concentration cell. In a concentration cell, both half-cells consist of the same electrode material and the same ions, but at different concentrations. The potential difference arises from the tendency of the ions to move from the more concentrated solution to the less concentrated solution to equalize the concentrations.

   The cell notation is: M(s) | M+(aq ; C₁) || M+(aq ; C₂) | M(s)

   The potential of a concentration cell is given by the Nernst equation: $E_{cell} = E_{cell}^\circ - \frac{2.303 RT}{nF} \log_{10} Q$ For a concentration cell, the standard cell potential $E_{cell}^\circ$ is zero because the electrodes are identical. The reaction quotient, Q, is the ratio of the ion concentration at the anode (oxidation, lower concentration) to that at the cathode (reduction, higher concentration). $Q = \frac{[M^+]_{anode}}{[M^+]_{cathode}} = \frac{C_1}{C_2}$ So, the Nernst equation simplifies to: $E_{cell} = - \frac{2.303 RT}{nF} \log_{10} \left( \frac{C_1}{C_2} \right) = \frac{2.303 RT}{nF} \log_{10} \left( \frac{C_2}{C_1} \right)$ Let the constant term $K = \frac{2.303 RT}{nF}$. The equation becomes: $E_{cell} = K \log_{10} \left( \frac{C_2}{C_1} \right)$

2. Analyze the first case.

   • Concentration at the anode, $C_1 = 0.05$ M.
   • Concentration at the cathode, $C_2 = 1$ M.
   • The magnitude of the cell potential, $|E_{cell,1}| = 70$ mV.

   Substituting these values into our simplified Nernst equation: $70 \text{ mV} = K \log_{10} \left( \frac{1}{0.05} \right)$ $70 = K \log_{10}(20) \quad \ldots(1)$

3. Analyze the second case.

   • The 0.05 M solution is replaced by a 0.0025 M solution. So, the new anode concentration is $C'_1 = 0.0025$ M.
   • The cathode concentration remains the same, $C_2 = 1$ M.
   • We need to find the new magnitude of the cell potential, $|E_{cell,2}|$.

   Using the same equation with the new concentration: $|E_{cell,2}| = K \log_{10} \left( \frac{1}{0.0025} \right)$ $|E_{cell,2}| = K \log_{10}(400) \quad \ldots(2)$

4. Solve for the new cell potential. We have a system of two equations:

   1. $70 = K \log_{10}(20)$
   2. $|E_{cell,2}| = K \log_{10}(400)$

   We can find the ratio of the two potentials by dividing equation (2) by equation (1): $\frac{|E_{cell,2}|}{70} = \frac{K \log_{10}(400)}{K \log_{10}(20)} = \frac{\log_{10}(400)}{\log_{10}(20)}$

   Now, we can simplify the logarithm term. Notice that $400 = 20^2$. $\log_{10}(400) = \log_{10}(20^2) = 2 \log_{10}(20)$

   Substitute this back into the ratio: $\frac{|E_{cell,2}|}{70} = \frac{2 \log_{10}(20)}{\log_{10}(20)} = 2$

   Therefore, the new cell potential is: $|E_{cell,2}| = 2 \times 70 \text{ mV} = 140 \text{ mV}$

5. Conclusion. When the concentration of the anode solution is changed from 0.05 M to 0.0025 M, the magnitude of the cell potential becomes 140 mV.