I have some difficulties to apply the basic concepts of calculating whether a reaction is spontanuous or not.


$$\ce{H_2SO_3(aq) + 2Mn(s) +4H^+(aq) -> S(s) + 2Mn^{2+}(aq) +3H_2 O(l)}$$

and those tables containing the stadnard reduction potentials:

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The formula to apply is the following: $E_{cell} = E(reduction)_{cathode} - E(oxidation)_{anode} $

$E_{cell} = 0.20 - (+1.23)$ $E_{cell} = -1.03$

The answer is that the reaction occurs sopontaniously. However this is not what I calculated, according to my calculations it doesn't. What am I doing wrong?

  • $\begingroup$ This 0.20 is for completely different reaction... $\endgroup$ – Mithoron Aug 19 '16 at 15:03
  • $\begingroup$ @Mithoron in that case could you tell me which value I should pick? I though 0.20 was correct. $\endgroup$ – privetDruzia Aug 19 '16 at 15:45
  • 1
    $\begingroup$ The correct reduction potential isn't on the table you provided. For the reduction of sulfurous acid to Sulfur, the reduction potential is +0.45 V. You also used the wrong reduction potential for the oxidation half reaction as well. The reduction potential for the Manganese(II) ion to solid Manganese is -1.18 V. $\endgroup$ – KeatonB Aug 19 '16 at 19:00

You need to approch the question in a systematic manner. The first step is to write the reduction and oxidation half reactions:

Reduction Half Reaction: (cathode)

$\ce{H2SO3 + 4 (H+) + 4 e- -> S(s) + 3 H2O}$

Oxidation Half Reaction: (anode)

$\ce{Mn -> (Mn^2+) + 2 e-}$

Now find the reduction potentials for both half reactions (only one was on the tables you provided, but a different set of tables might have both). Remember they are reduction half potentials so to find the reduction potential of the oxidation half reaction you must reverse the reaction direction:

$\ce{(Mn^2+) + 2 e- -> Mn}$

The reduction happens at the cathode, and oxidation happens at the anode.

$$\ce{E_{cell} = E_{cat} - E_{an}}$$

$\ce{E_{cat} = +0.45 V}$

$\ce{E_{an} = -1.18 V}$

$\ce{E_{cell} = (+0.45) - (-1.18)}$

$\ce{E_{cell} = +1.63 V (spontaneous)}$

Just to clarify again, the cathode's reduction potential (+0.45) was NOT on the tables you provided. I had to search the internet for a different set of tables which did have the cathode's half reaction reduction potential.


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