Consider the following redox reaction in basi | vedclass

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(B) The balanced chemical equation is obtained by splitting the reaction into two half-reactions:Oxidation half-reaction: $Cr(OH)_3 + 5(OH)^{-} \right

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$a + b = 7$

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(B) The balanced chemical equation is obtained by splitting the reaction into two half-reactions:Oxidation half-reaction: $Cr(OH)_3 + 5(OH)^{-} ightarrow (CrO_4)^{2-} + 4(H_2O) + 3e^{-}$Reduction half-reaction: $(IO_3)^{-} + 3(H_2O) + 6e^{-} ightarrow (I)^{-} + 6(OH)^{-}$To balance the electrons,multiply the oxidation half-reaction by $2$ and add it to the reduction half-reaction:$2Cr(OH)_3 + 10(OH)^{-} + (IO_3)^{-} + 3(H_2O) ightarrow 2(CrO_4)^{2-} + 8(H_2O) + (I)^{-} + 6(OH)^{-}$Simplifying the equation,we get:$2Cr(OH)_3 + (IO_3)^{-} + 4(OH)^{-} ightarrow 2(CrO_4)^{2-} + (I)^{-} + 5(H_2O)$Comparing with the given equation,we have $x=2, y=1, z=4, a=2, b=1, c=5$.Checking the options:$A: x + y = 2 + 1 = 3$ (Correct)$B: a + b = 2 + 1 = 3$ (Incorrect,as $3 
eq 7$)$C: z = 4$ (Correct)$D: b = 1$ (Correct)Therefore,the incorrect option is $B$.

Consider the following redox reaction in basic medium.
$x \ Cr(OH)_3 + y(IO_3)^{-} + z(OH)^{-} \rightarrow a(CrO_4)^{2-} + b(I)^{-} + c(H_2O)$
The incorrect option about it is

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Consider the following redox reaction in basic medium.$x \ Cr(OH)_3 + y(IO_3)^{-} + z(OH)^{-} \rightarrow a(CrO_4)^{2-} + b(I)^{-} + c(H_2O)$The incorrect option about it is