Question

Which of the following aqueous solutions remain alkaline after electrolysis ?

• A. $C{H}_{3}COONa$
• B. $KN{O}_3$
• C. $NaCl$
• D. $LiF$

Answer 1

Answer: (A), (C)

The correct options are
A $C{H}_{3}COONa$
C $NaCl$

a. Reduction potential of $H_{2}O>$ Reduction potential of $N{a}^{\oplus }$
Hence,
Cathode : $2H_{2}O+2e^{-}\to 2\stackrel{\ominus }{O}H+\begin{array}{c}{H}_2\\ \left(solutionisbasic\right)\end{array}$
Anode : $C{H}_{3}CO{O}^{\ominus }\stackrel{Kolb{e}^{\prime }selectrolysis}{\to }{C}_2{H}_6\left(Ethane\right)+2C{O}_2$

b. $K^{\oplus }$ does not undergo reduction but reduction of $H_{2}O$ occurs to give $\stackrel{\ominus }{O}H$ ion and $H_2\left(g\right)$.
Similarly, $N{O}_3^{\ominus }$ ion does not undergo oxidation but oxidation of $H_{2}O$ occurs to give $H^{\oplus }$ ions and $O_2\left(g\right)$.
$H^{\oplus }$ and $\stackrel{\ominus }{O}H$ ions get neutralised and pH = 7 (neutral solution).

c. $N{a}^{\oplus }$ ions do not undergo reduction but reduction of $H_{2}O$ occurs to give $\stackrel{\ominus }{O}H$ ions and $H_2\left(g\right)$. (Hence, solution is basic). $C{l}^{\ominus }$ undergoes oxidation to give $C{l}_2\left(g\right)$.

d. Same explanation as in (b).

Hence, options A and C are correct.