Question

Balance the following half reactions, in acidic conditions: Mn2O3(s) + H+(aq) + H2O(l) + e− → Mn2+(aq) + H+(aq) + H2O(l) + e− HAsO2(aq) + H+(aq) + H2O(l) + e− → As(s) + H+(aq) + H2O(l) + e−

Answer 1

I'd start by writing each half reaction (without the extra H's or electrons)

\[Mn_2O_3(s) \rightarrow Mn^{+2}(aq)\]and\[HAsO_2(aq) \rightarrow As(s)\]Make sure that all atoms \(other than\) oxygen and hydrogen are balanced first:
\[Mn_2O_3(s) \rightarrow 2Mn^{+2}(aq)\]\[HAsO_2(aq) \rightarrow As(s)\]Then balance the oxygen atoms by adding water to the side that's missing O's
\[Mn_2O_3(s) \rightarrow 2Mn^{+2}(aq) + 3H_2O(l)\]\[HAsO_2(aq) \rightarrow As(s) + 2H_2O(l)\]Balance the hydrogens by adding H+ ions to the necessary side
\[Mn_2O_3(s) + 6H^{+1}(aq) \rightarrow 2Mn^{+2}(aq) + 3H_2O(l)\]\[HAsO_2(aq) + 3H^{+1}(aq) \rightarrow As(s) + 2H_2O(l)\]Balance charge by adding electrons to one side, remembering that electrons are negative
\[Mn_2O_3(s) + 6H^{+1}(aq) + 2e^{-1} \rightarrow 2Mn^{+2}(aq) + 3H_2O(l)\]\[HAsO_2(aq) + 3H^{+1}(aq) + 3e^{-1} \rightarrow As(s) + 2H_2O(l)\]and now each half-reaction is balanced for mass and charge.

Answer 2

This is superb! Thanks a million!

Answer 3

yvw