Reaction of sulfate ion in copper sulfate electrolysis
A typical solution for copper plating contains about $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{CuSO4}$ and $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{H2SO4}$:

$$\begin{align}
c(\ce{Cu^2+})&amp;=0.5\ \mathrm{mol\ l^{-1}}\\
c(\ce{H+})&amp;=1\ \mathrm{mol\ l^{-1}}\\
c(\ce{SO4^2-})&amp;=1\ \mathrm{mol\ l^{-1}}\\
\end{align}$$

The positive ions $\ce{Cu^2+}$ and $\ce{H3O+}$ are attracted to the cathode. The redox potentials $E$ for $\mathrm{pH} = 0$ show that cathodic reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ is preferred over reduction of $\ce{H+}$ to $\ce{H2}$:

$$\begin{alignat}{2}
\ce{Cu^2+ + 2e- \;&amp;&lt;=&gt; Cu}\quad\quad &amp;&amp;E^\circ = +0.340\ \mathrm{V}\quad\quad &amp;&amp;&amp;E=E^\circ+\frac{0.059\ \mathrm{V}}{2}\log 0.5 = 0.331\ \mathrm{V}\\
\ce{2H+ + 2e- \;&amp;&lt;=&gt; H2}\quad\quad &amp;&amp;E^\circ = +0.000
\end{alignat}$$

At the anode, the redox potentials $E$ imply that oxidation of $\ce{H2O}$ to $\ce{O2}$ is preferred over oxidation of $\ce{SO4^2-}$:

$$\begin{alignat}{3}
\ce{O2 + 4H+ + 4e- \;&amp;&lt;=&gt; 2H2O}\quad\quad &amp;&amp;E^\circ = +1.229\ \mathrm{V}\\
\ce{S2O8^2- + 2e- \;&amp;&lt;=&gt; 2SO4^2-}\quad\quad &amp;&amp;E^\circ = +2.01\ \mathrm{V}
\end{alignat}$$

Nevertheless, $\ce{SO4^2-}$ actually is oxidised at the anode:
$$\ce{SO4^2- -&gt; SO4- + e-}$$
The reason for this reaction is that negative ions are attracted to the anode. Under neutral conditions, $\ce{OH-}$ ions would migrate to the anode where they would be oxidised to $\ce{O2}$. However, under the given acidic conditions, the concentration of $\ce{OH-}$ is very low. The predominant negative ions are $\ce{SO4^2-}$, which migrate to the anode.
Since the produced concentration of $\ce{SO4-}$ is low, the $\ce{SO4-}$ ions do not combine to $\ce{S2O8^2-}$. Instead $\ce{SO4-}$ reacts with water: 
$$\ce{4SO4- + 2H2O -&gt; 4SO4^2- + O2 + 4H+}$$

Therefore, the net reaction is the above-mentioned oxidation of $\ce{H2O}$ to $\ce{O2}$. The $\ce{SO4-}$ ions act as catalyst:

$$\begin{align}
\ce{4SO4^2- &amp;-&gt; 4SO4- + 4e-}\\
\ce{4SO4- + 2H2O &amp;-&gt; 4SO4^2- + O2 + 4H+}\\
\hline\ce{2H2O &amp;-&gt; O2 + 4H+ + 4e-}
\end{align}
$$