Question

What is the coefficient of H2O in the balanced half reaction? (See picture)

Answer 1

Firstly we would determine the oxidation number of the elements to determine which is undergoing oxidation and which is undergoing reduction:

`\begin{gathered} Mn\text{ }in\text{ }MnO_4^-:Mn+(-2*4)=-1 \\ Mn\text{ }in\text{ }MnO_4^-:Mn-8=-1 \\ Mn\text{ }in\text{ }MnO_4^-:Mn=-1+8=+7 \\ \\ Mn\text{ }in\text{ }MnO_4^(2-):Mn+(-2*4)=-2 \\ Mn\text{ }in\text{ }MnO_4^(2-):Mn=-2+8=+6 \end{gathered}`

Mn oxidation number goes from +7 to +6 which means it is undergoing a reduction. A decrease in oxidation number means reduction.

`\begin{gathered} S\text{ }in\text{ }HSO_3^-:1+(-2*3)+S=-1 \\ S\text{ }in\text{ }HSO_3^-:1-6+S=-1 \\ S\text{ }in\text{ }HSO_3^-:S=-1+5=+4 \\ \\ S\text{ }in\text{ }SO_4^(2-):S+(-2*4)=-2 \\ S\text{ }in\text{ }SO_4^(2-):S-8=-2 \\ S\text{ }in\text{ }SO_4^(2-):S=+6 \end{gathered}`

S oxidation goes from +4 to +6 meaning it is undergoing oxidation. There is an increase in oxidation number.

We will now balance the oxidation half reaction:

`HSO_3^-+H_2O\rightarrow SO_4^(2-)+3H^++2e`

The coefficient for the water molecule is 1 in the balnced half reaction.