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  4. Number of moles of K2Cr2O7 reduced by one mole of Sn2+ ions is

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Q. Number of moles of $K_2Cr_2O_7$ reduced by one mole of $Sn^{2+}$ ions is

Redox Reactions

Solution:

$Cr_2O^{2-}_7 + 14H^{+} + 3Sn^{2+} \rightarrow 2Cr^{3+} + 3 Sn^{4+} + 7 H_2O$
Thus, 1 mole of $Sn^{2+}$ reduces $\frac{1}{3}$ moles of $K_2Cr_2O_7$