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EXPERIMENT 1: Identify the oxidation and reduction half-reactions that occur in
Cell 5: Mn(s) | Mn(NO3)2 (aq) || Zn(NO3)2(aq) | Zn(s)
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Answer :

ANFABBA15GO

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

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