Question

One beaker contains 0.15 M $Cd(N{O}_3{)}_2$ and a Cd metal electrode. The other beaker contains 0.20 M $AgN{O}_3$ and a Ag metal electrode. The cell representation of a cell in which cadmium metal oxidizes

• A. Cd(s) | $C{d}^{2+}$ (aq, 0.15 M) || $A{g}^{+}$ (aq, 0.20 M) | Ag(s)
• B. $A{g}^{+}$ (aq, 0.20 M) | Ag(s) || Cd(s) | $C{d}^{2+}$ (aq, 0.15 M)
• C. $C{d}^{2+}$ | Cd(s) (aq, 0.15 M) || Ag(s) | $A{g}^{+}$ (aq, 0.20 M)

Answer 1

The correct option is A Cd(s) | $C{d}^{2+}$ (aq, 0.15 M) || $A{g}^{+}$ (aq, 0.20 M) | Ag(s)

As it is given cadmium metal is getting oxidized, so it is anode.
The net ionic equation for the cell reaction is:
$2A{g}^{+}\left(aq\right)+Cd\left(s\right)\to C{d}^{2+}\left(aq\right)+2Ag\left(s\right)$
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: $Cd\left(s\right)\to C{d}^{2+}\left(aq\right)+2e^{-}\left(oxidation\right)$
Cathode: $2A{g}^{+}\left(aq\right)+2e^{-}\to 2Ag\left(s\right)\left(reduction\right)$
cell notation is:
$Cd\left(s\right)|C{d}^{2+}(aq,0.15M)||A{g}^{+}(aq,0.20M)|Ag\left(s\right)$