6L.7 C

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

6L.7 C

Postby BeylemZ-1B » Sat Feb 22, 2020 2:32 pm

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

Postby Katherine Wu 1H » Sun Feb 23, 2020 3:24 pm

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)3+2e-→2Ni(OH)2(s)+2OH^-(aq) then, upon addition,
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

Postby Jasmine 2C » Mon Mar 02, 2020 2:53 am

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-
2Ni(OH)3+2e-→2Ni(OH)2(s)+2OH^-(aq) then, upon addition,
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

How do you know to put Ni(s) as the conducting electrode? Also, what about the KOH(aq)? What's the purpose of that?


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