InterviewSolution
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InterviewSolution
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The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions isA. `2//3`B. `1//6`C. `1//3`D. `1` |
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Answer» Correct Answer - C `undersetulbar(Cr(2)O_(7)^(2-)+14H^(+)2Sn^(2+) rarr 3Sn^(4+)+2Cr^(3+)+7H_(2)O)(Cr_(2)O_(7)^(2-)+14H^(+)+underset((Sn^(2+) rarr Sn^(4+)+2e^(-))xx3)(6e^(-)to2Cr^(3+)+7H_(2)O)` It is clear form this equation that `3` moles of `Sn^(2+)` reduce one mole of `Cr_(2)O_(7)^(2-)`, hence `1` mol. Of `Sn^(2+)` will reduce `(1)/(3)`moles of `Cr_(2)O_(7)^(2-)`. |
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