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प्रश्न
Give reasons:
On the basis of E° values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl.
Account for the following:
On the basis of E° values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl.
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उत्तर
While Cl2 gas is released during the electrolysis of aqueous NaCl, O2 gas should be released at the anode based on Eº values. It results from the over-voltage phenomenon. It has been found through experimentation that the voltage needed for electrolysis is higher than what is determined using standard potentials. We refer to this necessary additional voltage as over-voltage. It is necessary because both half reactions have a slow rate of electron transport at the electrode-solution contact. The amount of overvoltage needed to produce oxygen is far more than that needed to form chlorine.
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संबंधित प्रश्न
The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10−3 S cm−1.
Define conductivity for the solution of an electrolyte.
Give reasons:
Conductivity of CH3COOH decreases on dilution.
When measuring conductivity with a conductivity cell, cell constant denoted by the symbol G* is given by (electrode area of cross-section equal to ‘A’ and separated by distance ‘l’)
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(i) `1/R l/A`
(ii) `G^∗/R`
(iii) Λm
(iv) `l/A`
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Reason: Concentration of ionic solution will change if DC source is used.
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The conductivity of 0.01 m NaCl solution is 0.00147 ohm–1cm–1. What happens to the conductivity if extra 100 ML is added to the above solution:-
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