Define reduction in terms of (i) electron transfer, (ii) change in oxidation number - Leaving Cert Chemistry - Question 11 - 2022

Reduction also involves a decrease in the oxidation number of an atom in a molecule. This occurs when an atom gains electrons, resulting in a lower positive charge or a more negative charge.

The oxidation number of hydrogen in most of its compounds is +1 due to its higher electronegativity compared to most other elements. In hydrides, however, hydrogen is bonded to more electropositive elements, which results in it having an oxidation number of -1.

In most compounds, the oxidation number of oxygen is -2. However, in the case of OF₂ (oxygen difluoride), the oxidation number of oxygen is +2 because fluorine, being more electronegative, takes the oxidation state of -1.

First, assign oxidation numbers:

• Mn²⁺: +2
• BiO₃⁻: +5 for Bi, -2 for each O (there are 3 O)
• H⁺: +1
• MnO₄⁻: +7 for Mn, -2 for each O (there are 4 O)
• Bi³⁺: +3
• H₂O: 0 (H: +1, O: -2)

Next, balance the half-reactions:

1. For the oxidation half-reaction:

2. For the reduction half-reaction:

Finally, combine and balance charges to get:

2 Mn²⁺ + 5 BiO₃⁻ + 14 H⁺ → 2 MnO₄⁻ + 5 Bi³⁺ + 7 H₂O

The reducing agent in the reaction is Mn²⁺, as it donates electrons to reduce BiO₃⁻ and itself is oxidized to MnO₄⁻.