How many mL of 0.3 M $K_{2} C r_{2} O_{7}$ required for complete oxidation of 5 mL of 0.2 M $S n C_{2} O_{4}$ solution in acidic medium. ( $K_{2} C r_{2} O_{7}$ oxidises $S n C_{2} O_{4}$ to $S n^{4 +} + C O_{2}$ )

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answer is 2.22.

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Detailed Solution

$S n C_{2} O_{4} \overset{}{\to} S n^{4 +} + C O_{2}$
v.f of $S n C_{2} O_{4} = 2 + 2 = 4$
Now milli equivalent of $K_{2} C r_{2} O_{7}$ = milli equivalent of
$S n C_{2} O_{4} 0.3 \times 6 \times V = 0.2 \times 4 \times 5 V = 2.22 m L$

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