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Question
In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?
Options
\[\ce{Na+ (aq) + e- -> Na (s); E^{Θ}_{cell} = - 2.71V}\]
\[\ce{2H2O (l) -> O2(g) + 4H+ (aq) + 4e^- ; E^{Θ}_{cell} = 1.23V}\]
\[\ce{H^+ (aq) + e^- -> 1/2 H2 (g); E^{Θ}_{cell} = 0.00V}\]
\[\ce{Cl^- (aq) -> 1/2 Cl2 (g) + e^- ; E^{Θ}_{cell} = 1.36V}\]
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Solution
\[\ce{Cl^- (aq) -> 1/2 Cl2 (g) + e^- ; E^{Θ}_{cell} = 1.36V}\]
Explanation:
During electrolysis of aqueous
\[\ce{NaCl -> Na^+ + Cl^-}\]
\[\ce{H2O -> H+ + OH-}\]
\[\ce{Na+ + e- -> Na (E^{Θ}_{cell} = - 2.71V)}\]
\[\ce{H^+ e- -> 1/2 H2 E^{Θ}_{cell} = 0.00V}\]
At cathode,
\[\ce{H2O + e- -> 1/2 H2 + OH-}\]
At anode, two reactions are possible.
\[\ce{Cl^{-} -> 1/2 Cl2 + e- ; E^{Θ}_{cell} = 1.36V}\]
\[\ce{2H2O -> O2 + 4H+ + 4e- ; E^{Θ}_{cell} = 1.23V}\]
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