Question 9.1 [2 marks]
Definition of reduction in terms of oxidation numbers: Reduction refers to the process in which the oxidation number of an element decreases, indicating that the element gains electrons. An increase in the electron density around an atom in a compound typically results in a more negative oxidation state.
Question 9.2 [2 marks]
Calculating the oxidation number of phosphorus in HPO₄²⁻:
- The oxidation state of hydrogen (H) is +1.
- The oxidation state of oxygen (O) is -2.
- Let the oxidation state of phosphorus (P) be \( x \).
The sum of the oxidation states in the ion must equal the charge of the ion: \[ x + 1 + 4(-2) = -2 \] This simplifies to: \[ x + 1 - 8 = -2 \] \[ x - 7 = -2 \] \[ x = +5 \] Therefore, the oxidation number of phosphorus in HPO₄²⁻ is +5.
Question 9.3.1 [2 marks]
Formula of the reducing agent in this reaction: The reducing agent is the species that is oxidized and donates electrons in the reaction. In this case, the reducing agent is H₂S.
Question 9.3.2 [2 marks]
Reduction half-reaction: In this reaction, the permanganate ion (MnO₄⁻) is reduced to Mn²⁺. The half-reaction is: \[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
Question 9.3.3 [2 marks]
Oxidation half-reaction: In this reaction, hydrogen sulfide (H₂S) is oxidized to sulfur (S). The half-reaction is: \[ \text{H}_2\text{S} \rightarrow \text{S} + 2 \text{H}^+ + 2 \text{e}^- \]
These reactions clearly separate the reduction and oxidation processes occurring in the overall redox reaction.