## 6L.7 C

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### 6L.7 C

Write half reactions and devise a galvanic cell to study each of the following reactions:

(C) Cd + Ni(OH)3 —> Cd(OH)2 + 2Ni(OH)2
I don’t understand how in the answers for the homework, the half reaction for cathode looks like :
Ni(OH)3 + e- —> Ni(OH)2 + OH-
and then the cell diagram for the right side says:
|| Ni(OH)3| Ni(OH)2 | Ni(s)

Where is the solid nickel coming from? I’m lost

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
Been upvoted: 2 times

### Re: 6L.7 C

Cd(OH)2(s)+2e- →Cd(s)+2OH^-(aq)
E°(anode)=-0.81V
Ni(OH)3(s)+e-→ Ni(OH)2(s)+OH^-(aq)
E°(cathode)=+0.49V
Reversing the anode rxn and multiplying the cathode rxn by 2 yields
Cd(s)+2OH^-(aq)→Cd(OH)2(s)+2e-
2Ni(OH)2(s)+Cd(s)→Cd(OH)2(s)+2Ni(OH)2(s)
overall cell E° = +1.30V
and Cd(s)|Cd(OH)2(s)|KOH(aq)||Ni(OH)3(s)|NI(OH)2(s)|Ni(s)
with Ni(s) used as the conducting electrode for the cathode

Jasmine 2C
Posts: 184
Joined: Wed Sep 18, 2019 12:18 am

### Re: 6L.7 C

Katherine Wu 1H wrote:Cd(OH)2(s)+2e- →Cd(s)+2OH^-(aq)
E°(anode)=-0.81V
Ni(OH)3(s)+e-→ Ni(OH)2(s)+OH^-(aq)
E°(cathode)=+0.49V
Reversing the anode rxn and multiplying the cathode rxn by 2 yields
Cd(s)+2OH^-(aq)→Cd(OH)2(s)+2e-