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Postby Shanzey » Wed Feb 26, 2020 11:54 am

For 6L.7c, the solution uses Ni(s) as the electrode for the cathode in the cell diagram, but Ni(s) is neither a reactant or product of the reduction half reaction. Why do we have to use Ni(s)? Is it because Ni is already present in the solution/reaction and can act as a metal conductor? Would using Pt(s) or C(graphite) be incorrect?

Caitlyn Tran 2E
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Re: 6L.7

Postby Caitlyn Tran 2E » Wed Feb 26, 2020 12:37 pm

I think they just used nickel because it's a nickel-cadmium cell, so I assumed that those would be the electrodes. However, the sole purpose is to use an inert metal, so I think that using platinum or graphite would also be alright. If you want to double check, maybe try asking Professor Lavelle or a TA how this would be approached on a test? Hope this helps!

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Re: 6L.7

Postby 705302428 » Wed Feb 26, 2020 12:38 pm

I think platinum would work as well, like we did in class.

Ariel Davydov 1C
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Re: 6L.7

Postby Ariel Davydov 1C » Wed Feb 26, 2020 12:38 pm

Since Ni is present in its charged aqueous forms, we can use solid nickel as a spectator that allows for current to flow. I believe we can still use C(graphite) or Pt(s) as the electrode.

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