In the reactionx BrO_3^- + y Cr^{3+} + z H_2O | vedclass

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(D) The given reaction is: $x BrO_3^- + y Cr^{3+} + z H_2O 	o Br_2 + CrO_4^{2-} + H^{+}$Step $1$: Identify oxidation and reduction half-reactions.Red

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$6, 10, 22$

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(D) The given reaction is: $x BrO_3^- + y Cr^{3+} + z H_2O 	o Br_2 + CrO_4^{2-} + H^{+}$Step $1$: Identify oxidation and reduction half-reactions.Reduction: $BrO_3^- 	o Br_2$ (Bromine oxidation state changes from $+5$ to $0$)Oxidation: $Cr^{3+} 	o CrO_4^{2-}$ (Chromium oxidation state changes from $+3$ to $+6$)Step $2$: Balance the half-reactions.Reduction: $2 BrO_3^- + 12 H^+ + 10 e^- 	o Br_2 + 6 H_2O$Oxidation: $Cr^{3+} + 4 H_2O 	o CrO_4^{2-} + 8 H^+ + 3 e^-$Step $3$: Equalize electrons by multiplying reduction by $3$ and oxidation by $10$.$6 BrO_3^- + 36 H^+ + 30 e^- 	o 3 Br_2 + 18 H_2O$$10 Cr^{3+} + 40 H_2O 	o 10 CrO_4^{2-} + 80 H^+ + 30 e^-$Step $4$: Add the two half-reactions.$6 BrO_3^- + 10 Cr^{3+} + 22 H_2O 	o 3 Br_2 + 10 CrO_4^{2-} + 44 H^{+}$Comparing with the given equation,$x = 6$,$y = 10$,and $z = 22$.

In the reaction
$x BrO_3^- + y Cr^{3+} + z H_2O \to Br_2 + CrO_4^{2-} + H^{+}$
the coefficients $x, y, z$ are

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In the reaction$x BrO_3^- + y Cr^{3+} + z H_2O \to Br_2 + CrO_4^{2-} + H^{+}$the coefficients $x, y, z$ are