H^{+} + Cr_2O_7^{2-} + SO_3^{2-} to Cr^{3+} + | vedclass

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(A) The unbalanced equation is: $H^{+} + Cr_2O_7^{2-} + SO_3^{2-} 	o Cr^{3+} + SO_4^{2-} + H_2O$. Step $1$: Identify oxidation and reduction half-rea

Vedclass · Accepted Answer

$8, 3$

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(A) The unbalanced equation is: $H^{+} + Cr_2O_7^{2-} + SO_3^{2-} 	o Cr^{3+} + SO_4^{2-} + H_2O$. Step $1$: Identify oxidation and reduction half-reactions. Reduction: $Cr_2O_7^{2-} 	o Cr^{3+}$. Oxidation: $SO_3^{2-} 	o SO_4^{2-}$. Step $2$: Balance atoms and charges. Reduction half: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} 	o 2Cr^{3+} + 7H_2O$. Oxidation half: $SO_3^{2-} + H_2O 	o SO_4^{2-} + 2H^{+} + 2e^{-}$. Step $3$: Multiply oxidation half by $3$ to balance electrons: $3SO_3^{2-} + 3H_2O 	o 3SO_4^{2-} + 6H^{+} + 6e^{-}$. Step $4$: Add the two half-reactions: $Cr_2O_7^{2-} + 14H^{+} + 3SO_3^{2-} + 3H_2O 	o 2Cr^{3+} + 7H_2O + 3SO_4^{2-} + 6H^{+}$. Step $5$: Simplify: $Cr_2O_7^{2-} + 8H^{+} + 3SO_3^{2-} 	o 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$. The coefficient of $H^{+}$ is $8$ and the coefficient of $SO_4^{2-}$ is $3$.

$H^{+} + Cr_2O_7^{2-} + SO_3^{2-} \to Cr^{3+} + SO_4^{2-} + H_2O$. In the balanced reaction,the coefficients of $H^{+}$ and $SO_4^{2-}$ will be respectively:

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