Question

A voltaic cell consists of a strip of cadmium metal in a solution of cd(no3)2 in one beaker, and in the other beaker a platinum electrode is immersed in a nacl solution, with cl2 gas bubbled around the electrode. a salt bridge connects the two beakers. part a which electrode serves as the anode, and which as the cathode? which electrode serves as the anode, and which as the cathode? cd serves as the cathode, pt as the anode. pt serves as the cathode, cd as the anode. request answer part b does the cd electrode gain or lose mass as the cell reaction proceeds? does the electrode gain or lose mass as the cell reaction proceeds? the cd electrode will gain mass as the cell reaction proceeds. the cd electrode will lose mass as the cell reaction proceeds. request answer part c write the equation for the overall cell reaction.

Answer 1

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.

Step-by-step 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)

Answer 2

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)