1.

The charge required for the reduction of `1 mol` `Cr_(2)O_(7)^(2-)` ions to `Cr^(3+)` isA. `96500C`B. `2xx96500C`C. `3xx96500C`D. `6xx96500C`

Answer» Correct Answer - d
`6e^(c-)+Cr_(2)O_(7)^(2-) rarr 2Cr^(3+)`
So charge `=6xxF=6xx96500C`


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