$C r_{2} O_{7}^{2 -} + S O_{3}^{2 -} \overset{H^{+}}{\to} C r^{3 +} + S O_{4}^{2 -}$ .

The number of $H^{+}$ ions required to balance the above equation are

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answer is A.

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Detailed Solution

${\overset{+ 6}{C r}}_{2} O_{7}^{2 -} + \overset{+ 4}{S} O_{3}^{2 -} \overset{H^{+}}{\to} C r^{3 +} + \overset{+ 6}{S} O_{4}^{2 -} Change in oxidation no . of Cr = 2 (3) = 6 Change in oxidation no . of S = 2 Cross multiply the oxidation states (in the ratio 6 : 2 = 3 : 1) {\overset{+ 6}{C r}}_{2} O_{7}^{2 -} + 3 \overset{+ 4}{S} O_{3}^{2 -} \overset{H^{+}}{\to} 2 C r^{3 +} + \overset{+ 6}{3 S} O_{4}^{2 -}$
${Cr}_{2} O_{7}^{2 -} + 3 {SO}_{3}^{2 -} + 8 H^{+} \to 2 {Cr}^{3 +} + 3 {SO}_{4}^{2 -} + 4 H_{2} O$