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(i) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Omega. What is the cell constant, if the conductivity of 0.001 M KCl solution at 298 K is 0.146 xx 10^(-3) S cm^(-1)? (ii) Predict the products of electrolysis in the followingA solution of H_2SO_(4)with platinum electrodes. |
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Answer» SOLUTION : (i) Cell constant = Conductivity (or specific conductivity) x Resistance `=0.146 xx 10^(-3) S cm^(-1) xx 1500 Omega = 0.219 cm^(-1)` (ii) At cathode: `2H^(+) (AQ) + 2e^(-) to H_(2)(G)` At anode: `2OH^(-)(aq) to O_(2)(g) + 2H^(+) (aq) + 4e^(-)` Thus, `H_(2)`is evolved at the cathode and `O_(2)`at the anode. |
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