6L.7 C cell diagram

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Christineg1G
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6L.7 C cell diagram

Postby Christineg1G » Mon Mar 02, 2020 8:48 am

For 6L.7 C where do KOH(aq) and Ni(s) come from ? The half-reactions look like this:
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

Cell Diagram: Cd(s)|Cd(OH)2(s)|KOH(aq)||Ni(OH)3(s)|NI(OH)2(s)|Ni(s)

Brian Tangsombatvisit 1C
Posts: 119
Joined: Sat Aug 17, 2019 12:15 am

Re: 6L.7 C cell diagram

Postby Brian Tangsombatvisit 1C » Mon Mar 02, 2020 2:57 pm

The KOH is just another way of representing the presence of OH- ions at the anode. KOH will fully dissociate into K+ and OH- ions, so the K+ is a spectator ion and does not affect the reaction at the anode. If you were doing this experiment in a lab you would probably add OH- ions to the cell in the form of KOH. Ni is used as a metal electrode for the cathode since it requires a pure metal electrode even though the species are both solids.


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