I am working on a problem that provides these two half-reactions:

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

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?

  • $\begingroup$ What you propose is what's expected in terms of thermodynamics, and the standard reduction potential is the way to realise this. However, you have the freedom to write the equation backwards if you like; it won't happen, but you can certainly write it and calculate E° (which will be negative). $\endgroup$ Feb 6, 2019 at 13:46
  • $\begingroup$ Furthermore, just because it is not spontaneous under standard conditions doesn't mean it can't be done. This is what electrolysis does: it reverses the direction of spontaneous change. So in theory, what your question is asking could be achieved by applying an EMF in the direction of non-spontaneous change and thereby force it to happen. This is similar to how in thermodynamics you can have a change that is non-spontaneous, but can become spontaneous at higher temperature. In electrolysis the energy for that change is supplied by an EMF instead of by heat. $\endgroup$
    – Rafael
    Feb 6, 2019 at 15:14
  • $\begingroup$ It's also concentration-dependent (which include pH-dependence because one of the half-reactions includes hydrogen ions). If you start out in the absence of products, it will go forward (if just a tiny bit). $\endgroup$
    – Karsten
    Feb 6, 2019 at 18:07

1 Answer 1


Redox reactions are reversible. Despite the forward reaction is not spontaneous (as indicated by the unfavourable potential), it can still proceed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.