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Q.

A current of 0.01A is passed through a solution of iridium bromide. The only reaction at the cathode is the deposition of iridium metal. After 3.0 hrs, 0.072 gram of ‘Ir’ is deposited. The oxidation state of iridium ion is (atomic weight Ir = 192, F = 96,000 C) (Round off to nearest integer)

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answer is 3.

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Detailed Solution

0.01×3×60×60=n×0.072192×96000

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