Question

What will be the balanced equation in acidic medium for the given reaction?

$Cr_{2}O_{7(aq)}^{2 -} + SO_{2(g)} \rightarrow Cr_{(aq)}^{3 +} + SO_{4(aq)}^{2 -}$

• A. $Cr_{2}O_{7(aq)}^{2 -} + 3SO_{2(g)} + 2H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{3 +} + 3SO_{4(aq)}^{2 -}$ $+ H_{2}O_{(l)}$
• B. $2Cr_{2}O_{7(aq)}^{2 -} + 3SO_{2(g)} + 4H_{(aq)}^{+} \rightarrow 4Cr_{(aq)}^{3 +} + 3SO_{4(aq)}^{2 -}$ $+ 2H_{2}O_{(l)}$
• C. $Cr_{2}O_{7(aq)}^{2 -} + 3SO_{2(g)} + 14H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{3 +} + 3SO_{4(aq)}^{2 -}$ $+ 7H_{2}O_{(l)}$
• D. $Cr_{2}O_{7(aq)}^{2 -} + 6SO_{2(g)} + 7H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{3 +} + 6SO_{4(aq)}^{2 -}$ $+ 7H_{2}O_{(l)}$

Answer 1

Answer: (A)

$Cr_{2}O_{7(aq)}^{2 -} + 3SO_{2(g)} + 2H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{3 +} + 3SO_{4(aq)}^{2 -}$

$+ H_{2}O_{(l)}$

Answer 2

: $Cr_{2}O_{7}^{2 -} + SO_{2} \rightarrow Cr^{3 +} + SO_{4}^{2 -}$ (in acidic solution)

Oxidation half equation:

$SO_{2} + 2H_{2}O \rightarrow SO_{4}^{2 -} + 4H^{+} + 2e^{-}$ …(i)

Reduction half equation:

$Cr_{2}O_{7}^{2 -} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3 +} + 7H_{2}O$ …(ii)

Multiplying eqn. (i) by 3 and adding to eqn. (ii) we get

$Cr_{2}O_{7}^{2 -} + 3SO_{2} + 2H^{+} \rightarrow 2Cr^{3 +} + 3SO_{4}^{2 -} + H_{2}O$