$\mathrm{Fe}{}^{2+}$ in $1\mathrm{M}$ $\mathrm{H}{}_2\mathrm{SO}{}_4$ is titrated with $\mathrm{Ce}{}^{4+}$. Formal potentials: $E^{\circ }(\mathrm{Fe}{}^{3+}/\mathrm{Fe}{}^{2+})=0.68\mathrm{V}$, $E^{\circ }(\mathrm{Ce}{}^{4+}/\mathrm{Ce}{}^{3+})=1.44\mathrm{V}$. The potential of the platinum indicator electrode at the equivalence point is

For a one-electron oxidant titrating a one-electron reductant, the equivalence point potential is the average of the two standard (formal) potentials: $E_{eq}=\frac{E_1^{\circ }+E_2^{\circ }}{2}=\frac{0.68+1.44}{2}=1.06\mathrm{V}$. This follows from adding the two Nernst equations and using $[\mathrm{Fe}{}^{3+}]=[\mathrm{Ce}{}^{3+}]$, $[\mathrm{Fe}{}^{2+}]=[\mathrm{Ce}{}^{4+}]$ at equivalence.