A cell has two copper electrodes. Each beaker holds 100 mL of solution…

Step 1. The ratio. $\log\text{ratio} = \dfrac{0.0120}{0.02955} = 0.4061$, so the ratio is $2.55$.

Step 2. The equation. $\dfrac{0.500 - x}{0.0100 + x} = 2.55$. This gives $0.500 - x = 0.0255 + 2.55x$, so $x = \dfrac{0.4745}{3.55} = 0.1337\ \text{mol/L}$.

Step 3. The concentrations. Dilute side: $0.0100 + 0.1337 = 0.144\ \text{mol/L}$. Other side: $0.500 - 0.1337 = 0.366\ \text{mol/L}$.

Step 4. The charge. The copper moved is $0.1337 \times 0.100 = 0.01337\ \text{mol}$. Each mole carries 2 mol of electrons: $Q = 2 \times 0.01337 \times 96500 = 2580\ \text{C}$.