How to calculate whether a redox reaction is spontanuous?

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


given:


$$\ce{H\_2SO\_3(aq) + 2Mn(s) +4H^+(aq) -> S(s) + 2Mn^{2+}(aq) +3H\_2 O(l)}$$


and those tables containing the stadnard reduction potentials:


[![enter image description here](https://i.stack.imgur.com/VgPix.png)](https://i.stack.imgur.com/VgPix.png)


[![enter image description here](https://i.stack.imgur.com/ptVpB.png)](https://i.stack.imgur.com/ptVpB.png)


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?




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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