Can a half-reaction be oxidation if it has the higher reduction potential? I am working on a problem that provides these two half-reactions:

$$
\begin{align}
\ce{Fe^3+ (aq) + e- &amp;-&gt; Fe^2+ (aq)} &amp;\quad E^\circ_\mathrm{red} &amp;= \pu{+0.77 V}\\
\ce{S2O6^2- (aq) + 4H^+ + 2e- &amp;-&gt; 2H2SO3 (aq)} &amp;\quad E^\circ_\mathrm{red} &amp;= \pu{+0.60 V}
\end{align}
$$

What is confusing me is the way the question is worded:


  Write the balanced chemical equation for the oxidation of $\ce{Fe^2+ (aq)}$ by $\ce{S2O6^2- (aq)}$. 


Isn't this backwards though? Because the $\ce{Fe}$ half reaction has the higher reduction potential, shouldn't it be reduced while the $\ce{S2O6^2- (aq)}$ half reaction reverses to become oxidation?
Redox reactions are reversible. Despite the forward reaction is not spontaneous (as indicated by the unfavourable potential), it can still proceed.