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Give electrolytic reaction of aqueous NaCl solution (concentrated) in inert electrode. |
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Answer» Solution :* Inert electrodes are those electrodes which does not take part in reaction. * Solution is of NaCl, and on IONIZATION of NaCl it gives `Na^(+) and Cl^(-)` ions. In which, `Na^(+)` ions and water are near cathode. Which gives following reactions. `Na_((AQ))^(+)+e^(-) to Na""E_(cell)^(Theta)=-2.71V` or `H^(+)+e^(-) to (1)/(2)H_(2)""E_(cell)^(Theta)=0.0V` * So, reaction with more `E^(Theta)` VALUE gives reduction reaction. so near cathode `H^(+)` of water get reduced `(H_(2)O to H^(+)+OH^(-))` get reduced. (i) `Na^(+) + (H^(+)+OH^(-))+e^(-) to (1)/(2)H_(2(G)) +(Na^(+)+OH^(-))_((aq))` * So, near anode `Cl^(-)` and water, so oxidation reaction is possible. (ii) `Cl^(-)to(1)/(2)Cl_(2)+e^(-)""E_(cell)^(Theta)=1.36V` (iii) `2H_(2)O_((l)) to O_(2(g)) +4H_((aq))^(+) ""E_(cell)^(Theta)=1.23V` * Oxidation reaction is possible with less `E^(Theta)` values, so oxidation reaction of water with less `E^(Theta)` value should occur but oxidation of `Cl^(-)` occurs due to over voltage of oxygen oxidation of `Cl^(-)` is carried out by reaction (ii). * Overall reaction=(i)+(ii) `Na_((aq))^(+)+underset(H_(2)O)ubrace((H^(+)+OH^(-)))+Cl_((aq))^(-) to (1)/(2)Cl_(2(g))+(1)/(2)H_(2(g))+(Na^(+)+OH^(-))_((aq))` * So, near anode `Cl_(2)` gas and near cathode `H_(2)` gas and NaOH is occured. |
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