For the balanced redox reaction xMnO_4^- + yC | vedclass

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(A) The half-reactions are:Reduction: $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$Oxidation: $C_2O_4^{2-} 	o 2CO_2 + 2e^-$To balance the electrons,mul

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$2, 5, 16$

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(A) The half-reactions are:Reduction: $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$Oxidation: $C_2O_4^{2-} 	o 2CO_2 + 2e^-$To balance the electrons,multiply the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:$2MnO_4^- + 16H^+ + 10e^- 	o 2Mn^{2+} + 8H_2O$$5C_2O_4^{2-} 	o 10CO_2 + 10e^-$Adding these gives the balanced equation:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ 	o 2Mn^{2+} + 10CO_2 + 8H_2O$Comparing this with the given equation,we get $x = 2, y = 5, z = 16$.

For the balanced redox reaction $xMnO_4^- + yC_2O_4^{2-} + zH^+ \to Mn^{2+} + CO_2 + H_2O$,the values of $x, y$ and $z$ are respectively:

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