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Number of moles of `K_2Cr_2O_7` reduced by one mole of `Sn^(2+)` ion isA. `1/3`B. 3C. `1/6`D. 6 |
Answer» Correct Answer - A `underset"1 M"(Cr_2O_7^(2-)) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_2O` `underset"1 M"(Sn^(2+)) to Sn^(4+) + 2e^-` 1 M of `Sn^(2+)` on oxidation gives `2e^-` `6e^-` reduce 1 M of `Cr_2O_7^(2-)` completely. Then `2e^-` will reduce `M/6xx2=M/6` i.e.`1/3` moles of `K_2Cr_2O_7` |
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