1.

A current of 0.0965 amp is passed for 1000 seconds through 50 mL of 0.1 M NaCl. If the only reactions are reduction of H_(2)O to H_(2) at the cathode and oxidation of Cl^(-) to Cl_(2) at the anode, what will be the average concentration of OH^(-) in th e final solution ?

Answer»

Solution :Mole of ELECTRIC charge `= (0.0965 XX 1000)/(96500) = 0.001 F`.
`therefore` equivalent of `OH^(-)` LIBERATED = 0.001.
Mole of `OH^(-)` liberated = 0.001.
`therefore` concentration of `OH^(-)` in mole per litre `= (0.001"( mole)")/(0.05 "(litre)")`
= 0.02 M


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