Question

Which of the following statements is/are true for the electrolysis of $aq.LiCl$ with $Ag$ electrodes?

• A. $Ag$ get dissolved to give $A{g}^{+}$ at anode
• B. $Ag$ is deposited at cathode
• C. $O_2$ is liberated at anode
• D. $C{l}_2$ is liberated at anode

Answer 1

Answer: (A), (B)

$Ag$ is deposited at cathode

When the electrolysis of $aq.LiCl$ is done with $Ag$ electrodes, there are three possible reaction to occur at both anode and cathode.
Here, the silver electrode used is a active electrode. It is involved in the cell reaction.
At cathode:
1. $L{i}^{+}\left(aq\right)+e^{-}\rightleftharpoons Li\left(s\right)$
$E_{L{i}^{+}/Li}^0=-3.04V$

2. $2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$
$E_{H_{2}O/H_2}^0=-0.83V$

3. $A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$$E_{A{g}^{+}/Ag}^0=0.80V$

The reaction with a higher reduction value of $E^0$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At cathode : $A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$

At anode:
1. $2C{l}^{-}\left(aq\right)\rightleftharpoons C{l}_2\left(g\right)+2e^{-}$
$E_{C{l}^{-}/C{l}_2}^0=-1.36V$

2. $2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}\left(aq\right)+O_2\left(g\right)+4e^{-}$
$E_{H_{2}O/O_2}^0=-1.23V$

3. $Ag\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+e^{-}$
$E_{Ag/A{g}^{+}}^0=-0.8V$

The reaction with a higher oxidation value of $E^0$ takes place at the anode. Therefore, silver is dissolved at anode.
At anode: $Ag\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+e^{-}$