On balancing the given redox reaction,aCr_2O_ | vedclass

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Question

(A) Using the ion-electron method:
Reduction half-reaction: $Cr_2O_7^{2-} + 14H^{+} + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Oxidation half-reaction: $(

Vedclass · Accepted Answer

$1, 3, 8$

Vedclass · Answer

(A) Using the ion-electron method:
Reduction half-reaction: $Cr_2O_7^{2-} + 14H^{+} + 6e^- ightarrow 2Cr^{3+} + 7H_2O$
Oxidation half-reaction: $(SO_3^{2-} + H_2O ightarrow SO_4^{2-} + 2H^{+} + 2e^-) 	imes 3$
Adding both half-reactions:
$Cr_2O_7^{2-} + 14H^{+} + 6e^- + 3SO_3^{2-} + 3H_2O ightarrow 2Cr^{3+} + 7H_2O + 3SO_4^{2-} + 6H^{+} + 6e^-$
Simplifying the equation:
$Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^{+} ightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$
Comparing this with the given equation $aCr_2O_7^{2-} + bSO_3^{2-} + cH^{+} ightarrow 2aCr^{3+} + bSO_4^{2-} + \frac{c}{2}H_2O$:
$a = 1, b = 3, c = 8$.

On balancing the given redox reaction,$aCr_2O_7^{2-} + bSO_3^{2-}{(aq)} + cH^{+}{(aq)}$ $\rightarrow 2aCr^{3+}{(aq)} + bSO_4^{2-}{(aq)} + \frac{c}{2}H_2O_{(l)}$,the coefficients $a$,$b$,and $c$ are found to be,respectively-

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