Respuesta :
Reduction potential values:
Cd²⁺(aq) + 2e⁻ → Cd(s) E⁰red = - 0.403 V
Cl₂(g) + 2 e⁻ → 2 Cl⁻ (aq) E⁰red = 1.359 V
First part:
Electrode with positive reduction value acts as cathode Cl₂/Cl⁻ and electrode with negative reduction value acts as anode Cd²⁺/Cd
Second part:
Cd electrode is anode where oxidation takes place thus Cd(s) is oxidized to Cd²⁺(aq) Hence, Cd electrode loses mass as the cell reaction proceeds.
Third part:
Cell reaction:
Cd(s) → Cd²⁺(aq) + 2e⁻
Cl₂(g) + 2e⁻ → 2 Cl⁻(aq)
Cd(s) + Cl₂(g) → Cd²⁺(aq) + 2Cl⁻(aq)
Cd²⁺(aq) + 2e⁻ → Cd(s) E⁰red = - 0.403 V
Cl₂(g) + 2 e⁻ → 2 Cl⁻ (aq) E⁰red = 1.359 V
First part:
Electrode with positive reduction value acts as cathode Cl₂/Cl⁻ and electrode with negative reduction value acts as anode Cd²⁺/Cd
Second part:
Cd electrode is anode where oxidation takes place thus Cd(s) is oxidized to Cd²⁺(aq) Hence, Cd electrode loses mass as the cell reaction proceeds.
Third part:
Cell reaction:
Cd(s) → Cd²⁺(aq) + 2e⁻
Cl₂(g) + 2e⁻ → 2 Cl⁻(aq)
Cd(s) + Cl₂(g) → Cd²⁺(aq) + 2Cl⁻(aq)
Answer: 1. Pt serves as the cathode and Cd as the anode
2. Cd electrode loses mass as the cell reaction proceeds.
3. The overall cell reaction is mentioned in the explanation column.
Explanation: 1. The electrode with positive reduction value serves as cathode, that is, Pt, while, the electrode with negative reduction value functions as anode, that is, Cd.
2. The oxidation takes place in the Cd electrode, which is anode, and it gets oxidized to Cd²⁺ (aq). Hence, Cd electrode loses mass as the cell reaction moves forward.
3. Cell reaction:
Cd(s) = Cd²⁺ (aq) + 2e⁻
Cl₂ (g) + 2e⁻ = 2Cl⁻ (aq)
Cd(s) + Cl₂(g) = Cd²⁺ (aq) + 2Cl⁻ (aq)