In the balanced reaction xBrO_3^- + yCr^{3+} | vedclass

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(B) The given redox reaction is: $xBrO_3^- + yCr^{3+} + zH_2O 	o aBr_2 + bHCrO_4^- + cH^+$. Step $1$: Identify oxidation and reduction half-reactions

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$6, 10, 22$

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(B) The given redox reaction is: $xBrO_3^- + yCr^{3+} + zH_2O 	o aBr_2 + bHCrO_4^- + cH^+$. Step $1$: Identify oxidation and reduction half-reactions. Reduction: $BrO_3^- 	o Br_2$ (Bromine changes from $+5$ to $0$). Oxidation: $Cr^{3+} 	o HCrO_4^-$ (Chromium changes from $+3$ to $+6$). Step $2$: Balance the half-reactions. Reduction: $2BrO_3^- + 12H^+ + 10e^- 	o Br_2 + 6H_2O$. Oxidation: $Cr^{3+} + 4H_2O 	o HCrO_4^- + 7H^+ + 3e^-$. Step $3$: Multiply to equalize electrons. Multiply reduction by $3$ and oxidation by $10$: $6BrO_3^- + 36H^+ + 30e^- 	o 3Br_2 + 18H_2O$. $10Cr^{3+} + 40H_2O 	o 10HCrO_4^- + 70H^+ + 30e^-$. Step $4$: Add the equations. $6BrO_3^- + 10Cr^{3+} + 22H_2O 	o 3Br_2 + 10HCrO_4^- + 34H^+$. Comparing with the given equation,$x = 6$,$y = 10$,and $z = 22$.

In the balanced reaction $xBrO_3^- + yCr^{3+} + zH_2O \to aBr_2 + bHCrO_4^- + cH^+$,the coefficients $x, y$ and $z$ are

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