Question

What will be the balanced equation in acidic medium for the given reaction?
$Cr_{2}O^{2-}_{7\left(aq\right)} + SO_{2\left(g\right)} \to Cr^{3+}_{\left(aq\right)} + SO^{2-}_{4\left(aq\right)}$

• A. $Cr_{2}O^{2-}_{7\left(aq\right)} + 3SO_{2\left(g\right)} + 2H^{+}_{\left(aq\right)} \to 2Cr^{3+}_{\left(aq\right)} + 3SO^{2-}_{4\left(aq\right)}+ H_{2}O_{\left(l\right)}$
• B. $2Cr_{2}O^{2-}_{7\left(aq\right)} + 3SO_{2\left(g\right)} + 4H^{+}_{\left(aq\right)} \to 4Cr^{3+}_{\left(aq\right)} + 3SO^{2-}_{4\left(aq\right)}+ 2H_{2}O_{\left(l\right)}$
• C. $Cr_{2}O^{2-}_{7\left(aq\right)} + 3SO_{2\left(g\right)} + 14H^{+}_{\left(aq\right)} \to 2Cr^{3+}_{\left(aq\right)} + 3SO^{2-}_{4\left(aq\right)}+ 7H_{2}O_{\left(l\right)}$
• D. $Cr_{2}O^{2-}_{7\left(aq\right)} + 6SO_{2\left(g\right)} + 7H^{+}_{\left(aq\right)} \to 2Cr^{3+}_{\left(aq\right)} + 6SO^{2-}_{4\left(aq\right)}+ 7H_{2}O_{\left(l\right)}$

Answer 1

Answer: (A)

$Cr_{2}O^{2-} + SO_{2} \to Cr^{3+} + SO_{4}^{2-}$
(in acidic solution)
Oxidation half equation:
$SO_{2} + 2H_{2}O -> SO_{4} + 4H^{+} + 2e^{-} \quad...\left(i\right)$
Reduction half equation:
$Cr_{7}^{2-}O^{2-} + 14H^{+} + 6e^{-} \to 2Cr^{3+} + 7H_{2}O\quad ... \left(ii\right)$
Multiplying eqn. $\left(i\right)$ by $3$ and adding to eqn. $\left(ii\right)$ we get
$Cr_{2}O_{7}^{2-} + 3SO_{2} + 2H^{+} \to 2Cr^{3+} + 3SO_{4}^{2-} + H_{2}O$