For the given redox reaction: a MnO_4^{-} + b | vedclass

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(B) To balance the redox reaction in basic medium: $1$. Oxidation half-reaction: $S_2 O_3^{2-} \longrightarrow 2 SO_4^{2-} + 8 e^-$$2$. Reduction half

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$8; 6$

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(B) To balance the redox reaction in basic medium: $1$. Oxidation half-reaction: $S_2 O_3^{2-} \longrightarrow 2 SO_4^{2-} + 8 e^-$$2$. Reduction half-reaction: $MnO_4^{-} + 2 H_2 O + 3 e^- \longrightarrow MnO_2 + 4 OH^-$$3$. To balance electrons,multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $8$:$3 S_2 O_3^{2-} \longrightarrow 6 SO_4^{2-} + 24 e^-$$8 MnO_4^{-} + 16 H_2 O + 24 e^- \longrightarrow 8 MnO_2 + 32 OH^-$$4$. Adding both equations:$8 MnO_4^{-} + 3 S_2 O_3^{2-} + 16 H_2 O \longrightarrow 8 MnO_2 + 6 SO_4^{2-} + 32 OH^-$Simplifying $H_2 O$ and $OH^-$ on both sides:$8 MnO_4^{-} + 3 S_2 O_3^{2-} + H_2 O \longrightarrow 8 MnO_2 + 6 SO_4^{2-} + 2 OH^-$Comparing with the given equation,$a = 8$ and $y = 6$.

For the given redox reaction: $a MnO_4^{-} + b S_2 O_3^{2-} + H_2 O \longrightarrow x MnO_2 + y SO_4^{2-} + z OH^-$,the values of $a$ and $y$ respectively are:

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