KCET · Chemistry · p Block Elements (Group 13 & 14)
In the electrolysis of aqueous sodium chloride solution, which of the half cell reaction will occur
at anode?
- A \( N a^{+}(a q)+e^{-} \rightarrow N a(s) ; E_{\text {cell }}^{\circ}=-2.71 \) volts
- B \( 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 e^{-} ; \mathrm{E}_{\text {cell }}^{\circ}=1.23 \) \$volts
- C \( H^{+}(a q)+e^{-} \rightarrow \frac{1}{2} H_{2} ; E_{\text {cell }}^{\circ}=0.00 \) volts
- D \( \mathrm{Cl}^{-}(a q) \rightarrow \frac{1}{2} \mathrm{Cl}_{2}+e^{-} ; E_{\text {cell }}^{\circ}=1.36 \) volts
Answer & Solution
Correct Answer
(D) \( \mathrm{Cl}^{-}(a q) \rightarrow \frac{1}{2} \mathrm{Cl}_{2}+e^{-} ; E_{\text {cell }}^{\circ}=1.36 \) volts
Step-by-step Solution
Detailed explanation
In water sodium chloride ionizes as
\( N a C l(a q) \rightleftharpoons N a^{+}(a q)+C l^{-}(a q) \)
water also dissociates into ions, though to very slight degree as,
\( \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \)
Thus, the aqueous solution of sodium chloride contains \( \mathrm{N} a^{+}, \mathrm{H}^{+}, \mathrm{OH}^{-} \)and \( \mathrm{Cl}^{-} \)ions. When electric current is passed
through the solution, \( \mathrm{Cl}_{2} \) gas evolved at anode and hydrogen is evolved at the cathode. The resulting solution contains
\( \mathrm{Na}^{+} \)and \( \mathrm{OH}^{-} \)ions. Oxidation of \( \mathrm{Cl}^{-} \)ions and \( \mathrm{H}_{2} \mathrm{O} \) is possible at anode but, standard electrode potential of \( \mathrm{Cl}^{-} \)is
more, therefore, \( \mathrm{Cl}^{-} \)will oxidized at anode.
\( N a C l(a q) \rightleftharpoons N a^{+}(a q)+C l^{-}(a q) \)
water also dissociates into ions, though to very slight degree as,
\( \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \)
Thus, the aqueous solution of sodium chloride contains \( \mathrm{N} a^{+}, \mathrm{H}^{+}, \mathrm{OH}^{-} \)and \( \mathrm{Cl}^{-} \)ions. When electric current is passed
through the solution, \( \mathrm{Cl}_{2} \) gas evolved at anode and hydrogen is evolved at the cathode. The resulting solution contains
\( \mathrm{Na}^{+} \)and \( \mathrm{OH}^{-} \)ions. Oxidation of \( \mathrm{Cl}^{-} \)ions and \( \mathrm{H}_{2} \mathrm{O} \) is possible at anode but, standard electrode potential of \( \mathrm{Cl}^{-} \)is
more, therefore, \( \mathrm{Cl}^{-} \)will oxidized at anode.
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