InterviewSolution
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InterviewSolution
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In electrolysis of `NaCl` when `Pt` electrode is taken `H_(2)` is liberated at cathode while `Hg` cathode it forms sodium amalgam becauseA. Hg is more inert than PtB. more voltage is required to reduce `H^(+)` at Hg than at PtC. Na is dissovled in Hg while it does not dissolved in PtD. concentration of `H^(+)` ion is larger when Pt electrode is taken |
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Answer» Correct Answer - B (b) Sodium Chloride in water dissociates as `NaCl iff Na^(+) +Cl^(-)` `H_(2)O iff H^(+) +OH^(-)` When electric current is passed through this solution using platinum eletrodes , `Na^(+) and H^(+)` move towards cathode whereas `Cl^(-) and OH^(-)` ions move towards anode. At cathode `H^(+)+e^(-) to H` `H+H to H_(2)` At anode `Cl^(-) to Cl+e^(-)` `Cl+Cl to Cl_(2)` If mercury is used as cathode, `H^(+)` ions are not discharged at mercury cathode because mercury has high hydrogen over voltage. `Na^(+)` ions are discharged at cathode in preference of `H^(+)` ions yielding sodium, which dissolves in mercury to form sodium amalgum. |
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