In the following balanced reaction,X,MnO_4^- | vedclass

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(A) The given reaction is a redox reaction involving the reduction of permanganate and oxidation of oxalate.Reduction half-reaction:$MnO_4^- + 8H^+ +

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$2, 5, 16$

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(A) The given reaction is a redox reaction involving the reduction of permanganate and oxidation of oxalate.Reduction half-reaction:$MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$ ... $(i)$Oxidation half-reaction:$C_2O_4^{2-} 	o 2CO_2 + 2e^-$ ... $(ii)$To balance the electrons,multiply equation $(i)$ by $2$ and equation $(ii)$ by $5$:$2MnO_4^- + 16H^+ + 10e^- 	o 2Mn^{2+} + 8H_2O$$5C_2O_4^{2-} 	o 10CO_2 + 10e^-$Adding these two equations gives:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ 	o 2Mn^{2+} + 10CO_2 + 8H_2O$Comparing this with the given equation $X\,MnO_4^- + Y\,C_2O_4^{2-} + Z\,H^+ ightleftharpoons X\,Mn^{2+} + 2Y\,CO_2 + \frac{Z}{2}\,H_2O$,we get:$X = 2$,$Y = 5$,$Z = 16$.

In the following balanced reaction,
$X\,MnO_4^- + Y\,C_2O_4^{2-} + Z\,H^+ \rightleftharpoons X\,Mn^{2+} + 2Y\,CO_2 + \frac{Z}{2}\,H_2O$
values of $X, Y$ and $Z$ respectively are

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In the following balanced reaction,$X\,MnO_4^- + Y\,C_2O_4^{2-} + Z\,H^+ \rightleftharpoons X\,Mn^{2+} + 2Y\,CO_2 + \frac{Z}{2}\,H_2O$values of $X, Y$ and $Z$ respectively are