How to calculate whether a redox reaction is spontanuous?
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)

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

Oxidation Half Reaction: (anode)

$\ce{Mn -&gt; (Mn^2+) + 2 e-}$

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:

$\ce{(Mn^2+) + 2 e- -&gt; Mn}$

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.