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"Products obtained near the electrodes by electrolysis depends on the electrodes"-explain by suitable examples. |
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Answer» Solution :(i) and (ii) both are possible. The reaction which has more `E^(Theta)` value gives reaction. HENCE , reaction (ii) occurs near CATHODE. So, `H^(+)` ions of water get reduced and produce `H_(2)` gas. * So, on electrolysis of equeous solution of NaCl, `H_(2)` gas is obtained near cathode and NaOH in the solution by following reaction. (iii) `H_(2)O_((L))+e^(-)to(1)/(2)H_(2(g))+OH_((aq))^(-)` * So reaction, (i)+(ii)=reaction (iii) and spectator ions are produced near to cathode by the following reactions. (iv) `Na_((aq))^(+)+H_(2)O_((l))+e^(-) to (1)/(2)H_(2(g))+OH_((aq))^(-)+Na^(+)` * Experimentally, pink colour obtained with phenolphthalein near cathode gives the confirmation of presence of NaOH solution. |
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