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6L.7 C

Posted: Sat Feb 22, 2020 2:32 pm
by BeylemZ-1B
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

Re: 6L.7 C

Posted: Sun Feb 23, 2020 3:24 pm
by Katherine Wu 1H
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

Re: 6L.7 C

Posted: Mon Mar 02, 2020 2:53 am
by Jasmine 2C
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?