Answer:
Cathode: Mn → Mn²⁺ + 2e⁻ (Oxidation)
Anode: Zn²⁺ + 2e⁻ → Zn (Reduction)
Mn | Mn²⁺ || Zn²⁺ | Zn
Explanation:
To identify the half reaction we need to see the oxidation states.
Mn(s) → Ground state → Oxidation state = 0
Mn(NO₃)₂ → Mn²⁺ → The oxidation state has increased.
This is the oxidation reaction. It has released two electrons:
Mn → Mn²⁺ + 2e⁻
Zn(NO₃)₂ → Zn²⁺
Zn → Ground state → The oxidation state was decreased.
This is the reduction reaction. It has gained two electrons:
Zn²⁺ + 2e⁻ → Zn
Cathode: Mn → Mn²⁺ + 2e⁻
Anode: Zn²⁺ + 2e⁻ → Zn
Mn | Mn²⁺ || Zn²⁺ | Zn
