

InterviewSolution
Saved Bookmarks
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` |
|