Question

which side will the electrons be added to in the half - reaction?
$8\ce{h+} + \ce{mno^-_{4}} \
ightarrow \ce{mn^{2+}} + 4\ce{h2o}$
a. add electrons to the reactant side
b. add electrons to the product side

Explanation:

To determine where electrons are added, we analyze the oxidation state of Mn. Mn in \(MnO_4^-\) has an oxidation state of +7, and in \(Mn^{2+}\) it is +2. A decrease in oxidation state (from +7 to +2) means Mn is reduced (gains electrons). In a reduction half - reaction, electrons are reactants (added to the reactant side) because the species being reduced gains electrons. So, for the reduction of \(MnO_4^-\) to \(Mn^{2+}\), electrons are added to the reactant side.

Answer:

A. Add electrons to the reactant side