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. If the reduction potential, $E_{A{g}^{+}/Ag}^0=0.77Vand{E}_{C{d}^{2+}/Cd}^0=-0.40$
The cell representation of a cell is:

• A. $Cd\left(s\right)|C{d}^{2+}(aq,0.15M)||A{g}^{+}(aq,0.20M)|Ag\left(s\right)$
• B. $A{g}^{+}(aq,0.20M)|Ag\left(s\right)||Cd\left(s\right)|C{d}^{2+}(aq,0.15M)$
• C. $C{d}^{2+}|Cd\left(s\right)(aq,0.15M)||Ag\left(s\right)|A{g}^{+}(aq,0.20M)$
• D. None of the above.

Answer 1

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

To get a positive value of the cell potential,
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:
$Cd\left(s\right)|C{d}^{2+}(aq,0.15M)||A{g}^{+}(aq,0.20M)|Ag\left(s\right)$