For the ionic reaction: XBrO_3^{-} + YCr^{+3} | vedclass

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(C) The balanced redox reaction is determined by balancing the oxidation and reduction half-reactions.$BrO_3^{-} \rightarrow Br_2$ (Reduction: $Br^{+5

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$X = 6, Y = 10, Z = 22$

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(C) The balanced redox reaction is determined by balancing the oxidation and reduction half-reactions.$BrO_3^{-} \rightarrow Br_2$ (Reduction: $Br^{+5}$ to $Br^0$,change of $5$ per $Br$,so $10$ per $Br_2$)$Cr^{+3} \rightarrow HCrO_4^{-}$ (Oxidation: $Cr^{+3}$ to $Cr^{+6}$,change of $3$ per $Cr$)To balance the electrons,multiply the reduction half by $3$ and the oxidation half by $10$ (or simplify to $3$ and $5$ for $Br_2$ and $Cr$ respectively).$6BrO_3^{-} + 10Cr^{+3} + 22H_2O \rightarrow 3Br_2 + 10HCrO_4^{-} + 34H^{+}$Comparing coefficients,we get $X = 6, Y = 10, Z = 22$.

For the ionic reaction: $XBrO_3^{-} + YCr^{+3} + ZH_2O \rightarrow pBr_2 + qHCrO_4^{-} + rH^{+}$

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