Respuesta :
Reduction potential values:
CdÂČâș(aq) + 2eâ»Â â Cd(s)   Eâ°red = - 0.403 V
Clâ(g) + 2 eâ»Â â 2 Clâ» (aq)  Eâ°red = 1.359 V
First part:
Electrode with positive reduction value acts as cathode Clâ/Clâ» and electrode with negative reduction value acts as anode CdÂČâș/Cd
Second part:
Cd electrode is anode where oxidation takes place thus Cd(s) is oxidized to CdÂČâș(aq) Hence, Cd electrode loses mass as the cell reaction proceeds.
Third part:
Cell reaction:
Cd(s)Â â CdÂČâș(aq) + 2eâ»
Clâ(g) + 2eâ»Â â 2 Clâ»(aq)
Cd(s) + Clâ(g)Â â CdÂČâș(aq) + 2Clâ»(aq)
CdÂČâș(aq) + 2eâ»Â â Cd(s)   Eâ°red = - 0.403 V
Clâ(g) + 2 eâ»Â â 2 Clâ» (aq)  Eâ°red = 1.359 V
First part:
Electrode with positive reduction value acts as cathode Clâ/Clâ» and electrode with negative reduction value acts as anode CdÂČâș/Cd
Second part:
Cd electrode is anode where oxidation takes place thus Cd(s) is oxidized to CdÂČâș(aq) Hence, Cd electrode loses mass as the cell reaction proceeds.
Third part:
Cell reaction:
Cd(s)Â â CdÂČâș(aq) + 2eâ»
Clâ(g) + 2eâ»Â â 2 Clâ»(aq)
Cd(s) + Clâ(g)Â â CdÂČâș(aq) + 2Clâ»(aq)
Answer: 1. Pt serves as the cathode and Cd as the anode
2. Cd electrode loses mass as the cell reaction proceeds. Â
3. The overall cell reaction is mentioned in the explanation column. Â
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. Â
2. The oxidation takes place in the Cd electrode, which is anode, and it gets oxidized to CdÂČâș (aq). Hence, Cd electrode loses mass as the cell reaction moves forward. Â
3. Cell reaction: Â
Cd(s) = CdÂČâș (aq) + 2eâ»
Clâ (g) + 2eâ» = 2Clâ» (aq)
Cd(s) + Clâ(g) = CdÂČâș (aq) + 2Clâ» (aq)