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Question

One beaker contains 0.15 M Cd(NO3)2 and a Cd metal electrode. The other beaker contains 0.20 M AgNO3 and a Ag metal electrode. If the reduction potential, E0Ag+/Ag=0.77V and E0Cd2+/Cd=0.40
The cell representation of a cell is:

A
Cd(s)|Cd2+(aq, 0.15 M) || Ag+(aq, 0.20 M)|Ag(s)
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B
Ag+(aq, 0.20 M)|Ag(s) || Cd(s)|Cd2+(aq, 0.15 M)
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C
Cd2+|Cd(s)(aq, 0.15 M) || Ag(s)|Ag+(aq, 0.20 M)
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D
None of the above.
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Solution

The correct option is A Cd(s)|Cd2+(aq, 0.15 M) || Ag+(aq, 0.20 M)|Ag(s)
To get a positive value of the cell potential,
The net ionic equation for the cell reaction is:
2 Ag+(aq)+Cd(s)Cd2+(aq)+2Ag(s)
In the reaction above, cadmium metal is oxidized (loses electrons) and is, therefore, the anode. The silver ion is reduced (gains electrons) to form Ag (the cathode).
Anode: Cd(s)Cd2+(aq)+2 e(oxidation)
Cathode: 2 Ag+(aq)+2 e2 Ag(s)(reduction)
cell notation is:
Cd(s) | Cd2+(aq,0.15 M) || Ag+(aq, 0.20 M) | Ag(s)

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