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Explain why electrolysis of aqueous solution of NaCl gives H_(2)at cathode and Cl_(2) at anode. Write the overall reaction. [Given : E_(Na^(+)//Na)^(@) = -2.71 V, E_(Cl_(2)//2Cl^(-))^(@) = 1.36 V] 1/2 O_(2)(g) + 2H^(+) (aq) + 2e^(-) to H_(2)O (l), E^(@) = 1.23 V |
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Answer» Solution :The electrode reactions may be represented as under: `NaCl(AQ) to NA^(+) (aq) + Cl^(-) (aq)` `H_(2)O(l) to H^(+) (aq) + OH^(-)` (aq) At cathode: `2H^(+) (aq) + 2E^(-) to H_(2)` (g) It is because reduction potential of `E_(H^(+)//H_(2))^(@)` is greater than that of `E_(Na^(+)//Na)^(@)` At anode: `2Cl^(-)(aq) -2 E^(-) to Cl_(2) (g)` It is because of overvoltage i.e., energy required to liberate `O_(2)`is more than that required to liberate `Cl_(2)` Overall reaction: `2NaCl (aq) + 2H_(2)O (l) OVERSET("Electrolysis") to H_(2)(g) + Cl_(2) (g) + 2Na^(+) (aq) + 2OH^(-)` (aq) |
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