A solution holds 0.050 mol/L sodium iodide and 0.20 mol/L sodium bromide…

Step 1. The first. Iodine has the lower standard potential, so iodide is oxidised first: $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$.

Step 2. Bromine starts. $E = 1.09 - 0.0591 \times \log 0.20 = 1.09 + 0.0413 = 1.131\ \text{V}$.

Step 3. The iodide left. $0.54 - 0.0591\log[\text{I}^-] = 1.131$, so $\log[\text{I}^-] = \dfrac{0.54 - 1.131}{0.0591} = -10.0$ and $[\text{I}^-] = 9.9 \times 10^{-11}\ \text{mol/L}$. The share left is $\dfrac{9.9 \times 10^{-11}}{0.050} = 2.0 \times 10^{-9}$.

Step 4. The charge. The iodide oxidised is $0.999 \times 0.050 = 0.04995\ \text{mol}$ for each litre, and each ion gives one electron. $Q = 0.04995 \times 96500 = 4820\ \text{C}$.