Question

For a particular redox reaction SO32– is oxidized to SO42– and Fe3 is reduced to Fe2 . Complete and balance the equation for this reaction in basic solution. Phases are optional.

Answer 1

\[ SO_3^{2–} \rightarrow SO_4^{2–} \]Oxidation number(O.N.) change of S = (+6) - (+4)= 2

\[Fe^{3+} \rightarrow Fe^{2+}\]
O.N. change of Fe = (+2) - (+3) = -1

So, multiply the second equation by 2 and combine the two equations.
\[SO_3^{2–} + 2Fe^{3+}\rightarrow SO_4^{2–} +2Fe^{2+}\]
Then, balance the equation by adding \(H_2O\) and \(H^+\) to balance the number of O and H.
\[SO_3^{2–} + 2Fe^{3+} + H_2O\rightarrow SO_4^{2–} +2Fe^{2+}\]\[SO_3^{2–} + 2Fe^{3+} + H_2O\rightarrow SO_4^{2–} +2Fe^{2+} + 2H^+\]
Check if the charges on both sides are balanced. But it should be balanced :|