We are dealing with a redox titration problem involving the titration of Fe²⁺ by MnO₄⁻ in acidic conditions, specifically in 1 M H₂SO₄ at pH 0. The goal is to calculate the cell potential at specific volumes of titrant added: 13 mL, 15 mL, and 17 mL.

To solve this, we'll break it down into the following steps:

1. Identify the Redox Reaction: $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$ This is a 5:1 stoichiometric reaction between Fe²⁺ and MnO₄⁻.

2. Determine the Initial Number of Moles of Fe²⁺ and MnO₄⁻:

   • The initial concentration of Fe²⁺ is 3.75 mM and the volume is 400.0 mL. $\text{moles of Fe}^{2+} = 3.75 \times 10^{-3} \, \text{mol/L} \times 0.4000 \, \text{L} = 1.5 \times 10^{-3} \, \text{mol}$

   • The concentration of MnO₄⁻ is 20.0 mM, and we will calculate moles for each titrant volume.

     For $V_{\text{titran}} = 13 \, \text{mL}$: $\text{moles of MnO}_4^- = 20.0 \times 10^{-3} \, \text{mol/L} \times 0.013 \, \text{L} = 2.6 \times 10^{-4} \, \text{mol}$

     For $V_{\text{titran}} = 15 \, \text{mL}$: $\text{moles of MnO}_4^- = 20.0 \times 10^{-3} \, \text{mol/L} \times 0.015 \, \text{L} = 3.0 \times 10^{-4} \, \text{mol}$

     For $V_{\text{titran}} = 17 \, \text{mL}$: $\text{moles of MnO}_4^- = 20.0 \times 10^{-3} \, \text{mol/L} \times 0.017 \, \text{L} = 3.4 \times 10^{-4} \, \text{mol}$

3. Determine the Number of Moles of Fe²⁺ and Fe³⁺ Remaining: Based on the stoichiometric ratio, for every mole of MnO₄⁻, 5 moles of Fe²⁺ are oxidized.

   • For $V_{\text{titran}} = 13 \, \text{mL}$: $\text{moles of Fe}^{2+} \text{ consumed} = 5 \times 2.6 \times 10^{-4} = 1.3 \times 10^{-3} \, \text{mol}$ $\text{moles of Fe}^{2+} \text{ remaining} = 1.5 \times 10^{-3} - 1.3 \times 10^{-3} = 2.0 \times 10^{-4} \, \text{mol}$ $\text{moles of Fe}^{3+} \text{ formed} = 1.3 \times 10^{-3} \, \text{mol}$

   • For $V_{\text{titran}} = 15 \, \text{mL}$: [ \text{moles of Fe}^{2+} \text{ consumed} = 5 \times 3.0 \