Question

What is the coefficient of H2O in the balanced half reaction?

Answer 1

Step-by-step explanation

Given that:

What to find:

The coefficient of H2O in the balanced oxidation half-reaction.

Solution:

Step 1: Identify the species undergoing oxidation and reduction.

Mn changes from +7 in MnO₄⁻ to +6 in MnO₄²⁻. It is undergoing a reduction

S changes from +4 in HSO₃⁻ to +6 in SO₄²⁻. It is undergoing oxidation.

Step 2: Write the half-reaction for oxidation and reduction.

`\begin{gathered} Oxidation:HSO_3^-\rightarrow SO_4^(2-)+2e^- \\ \\ Reduction:MnO_4^-+e^-\rightarrow MnO_4^(2-) \end{gathered}`

Step 3: Balance the half-reactions by equating the number of electrons on both sides.

`\begin{gathered} Oxidation:(HSO_3^-\rightarrow SO_4^(2-)+2e^-)*1 \\ \\ Reduction:(MnO_4^-+e^-\rightarrow MnO_4^(2-))*2 \\ \\ Oxidation:HSO_3^-\rightarrow SO_4^(2-)+2e^- \\ \\ Reduction:2MnO_4^-+2e^-\rightarrow2MnO_4^(2-) \end{gathered}`

Step 4: The number of electrons will be canceled so the overall reaction will be:

`2MnO_4^-+HSO_3^-\rightarrow2MnO_4^(2-)+SO_4^(2-)`

Step 5: Balance the number of Oxygen atoms by adding H₂O on the left side.

`2MnO_4^-+HSO_3^-+H_2O\rightarrow2MnO_4^(2-)+SO_4^(2-)`

Step 6: Balance the number of H by adding H⁺ (acid) on the right side.