2 MnO_4^{-} + bI^{-} + cH_2O rightarrow xI_2 | vedclass

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(A) The given redox reaction is: $2 MnO_4^{-} + bI^{-} + cH_2O \rightarrow xI_2 + yMnO_2 + zOH^{-}$Step $1$: Write the half-reactions.Reduction half:

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$8$

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(A) The given redox reaction is: $2 MnO_4^{-} + bI^{-} + cH_2O \rightarrow xI_2 + yMnO_2 + zOH^{-}$Step $1$: Write the half-reactions.Reduction half: $MnO_4^{-} \rightarrow MnO_2$Balancing $O$ and $H$ in basic medium: $MnO_4^{-} + 2 H_2O + 3 e^{-} \rightarrow MnO_2 + 4 OH^{-}$Multiply by $2$: $2 MnO_4^{-} + 4 H_2O + 6 e^{-} \rightarrow 2 MnO_2 + 8 OH^{-}$Step $2$: Oxidation half.Oxidation half: $I^{-} \rightarrow I_2$Balancing atoms and electrons: $2 I^{-} \rightarrow I_2 + 2 e^{-}$Multiply by $3$: $6 I^{-} \rightarrow 3 I_2 + 6 e^{-}$Step $3$: Add the two half-reactions.$2 MnO_4^{-} + 6 I^{-} + 4 H_2O \rightarrow 2 MnO_2 + 3 I_2 + 8 OH^{-}$Comparing this with the given equation,the coefficient $z$ corresponds to the number of $OH^{-}$ ions,which is $8$.

Column-$I$	Column-$II$
$A$. $C_2O_4^{2-} \rightarrow 2CO_2$	$P$. $2$
$B$. $Cr_2O_7^{2-} + 14H^{+} \rightarrow 2Cr^{3+}$	$Q$. $6$
$C$. $2Na_2S_2O_3 \rightarrow Na_2S_4O_6$	$R$. $3$
$D$. $MnO_4^{-} \rightarrow MnO_2$	$S$. $1$

$2 MnO_4^{-} + bI^{-} + cH_2O \rightarrow xI_2 + yMnO_2 + zOH^{-}$
If the above equation is balanced with integer coefficients,the value of $z$ is . . . . . .

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$2 MnO_4^{-} + bI^{-} + cH_2O \rightarrow xI_2 + yMnO_2 + zOH^{-}$If the above equation is balanced with integer coefficients,the value of $z$ is . . . . . .