What is the balanced equation for the reaction of potassium permanganate, iron sulfate, and sulfuric acid?
Your first half reaction, for the reduction, is correct:
$$\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 -&gt; K2SO4 + 2 MnSO4 + 8 H2O}\tag1\label{red}$$

For the second half reaction, the oxidation, start by balancing iron:
\begin{align}
\ce{FeSO4 &amp;-&gt; Fe2(SO4)3 }\tag{2a}\\
\ce{2 FeSO4 &amp;-&gt; Fe2(SO4)3 }\tag{2b}
\end{align}

Add $\ce{H2SO4}$ on the left so you can balance sulfur:
\begin{align}
\ce{2 FeSO4 &amp;-&gt; Fe2(SO4)3 }\tag{2c}\\
\ce{H2SO4 + 2 FeSO4 &amp;-&gt; Fe2(SO4)3 }\tag{2d}
\end{align}

Now balance the protons and electrons:
\begin{align}
\ce{H2SO4 + 2 FeSO4 &amp;-&gt; Fe2(SO4)3 }\tag{2d}\\
\ce{H2SO4 + 2 FeSO4 &amp;-&gt; Fe2(SO4)3 + 2 H+ + 2 e-}\tag{2e}\label{ox}\\
\end{align}

Now add \eqref{red} and five times \eqref{ox} so that the electrons are equal on every side and coincidentally the protons also balance:
\begin{align}
\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 &amp;-&gt; K2SO4 + 2 MnSO4 + 8 H2O}\tag1\\
\ce{5 H2SO4 + 10 FeSO4 &amp;-&gt; 5 Fe2(SO4)3 + 10 H+ + 10 e-}\tag{$5\times$2e}
\end{align}

And the final result is:
\begin{align}
\ce{2 KMnO4 + 8 H2SO4 + 10 FeSO4 &amp;-&gt; K2SO4 + 2 MnSO4 + 8 H2O + 5 Fe2(SO4)3}\tag3
\end{align}