complete the half reactions for the cell shown here, and show the correct shorth

Question

complete the half reactions for the cell shown here, and show the correct shorthand notation for the cell by dragging the labels into the correct position. The electrode on the left is the anode, the one on the right is the cathode.

So there is only one beaker filled with the solution KOH(aq) there are two electrodes immeressed  in it. on the left there is the electrode Ni, Ni(OH)2 and on the right there is the electrode Cu, Cu(OH)2.

Anode half reaction:

Blank + 2OH^-equlibrium arrown yields Blank + 2e^-

Cathode Half Reaction:

Blank + 2e^- equlibrium arrow yields blank + 2OH^-

Please also provide the shorthand notation?  

Cu(S), Ni(s), KOH(aq), Ni(Oh)2(s), Cu(OH)2(s)

Explanation / Answer

Anode half-reaction: oxidation of Ni to Ni(OH)2

Two electrons are required to oxidize each Ni:

Ni(s) + 2 OH-(aq) => Ni(OH)2(s) + 2 e-


Cathode half reaction: reduction of Cu(OH)2 to Cu

Two electron are required to reduce each Cu(OH)2:

Cu(OH)2(s) + 2 e- => Cu(s) + 2 OH-(aq)


Cell notation:

(Note | is used to separate different phases)

Ni(s) | Ni(OH)2(s) | KOH(aq) | Cu(OH)2(s) | Cu(s)

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