Discussion
To balance the following reaction :
SO3(aq.)2- + H2O → SO4(aq.)2- + 2H+ (aq.)
one needs to :
Answer: B
One needs to add 2e- to the right side.
The atoms are already balanced: one S, four O and two H on each side. The charges are not: the left side carries -2 (the sulfite ion) and the right side -2 + 2(+1) = 0. Sulfur is oxidised from +4 in sulfite to +6 in sulfate, losing two electrons, so this is an oxidation half-reaction and the electrons appear as products:
\[ \begin{aligned} &\mathrm{SO_{3}^{2-} + H_{2}O} \\ &\rightarrow \mathrm{SO_{4}^{2-} + 2H^{+} + 2e^{-}} \end{aligned} \]
Now both sides carry -2. Adding electrons to the left would make the charge -4 there. Adding water would unbalance the atoms, and multiplying by 2 balances nothing.
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