MnO_4^{-} reacts with A^{x+} to form AO_3^{-} | vedclass

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(A) The redox reaction involves the reduction of $MnO_4^{-}$ to $Mn^{2+}$ and the oxidation of $A^{x+}$ to $AO_3^{-}$.For $MnO_4^{-} \rightarrow Mn^{2

Vedclass · Accepted Answer

$1$

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(A) The redox reaction involves the reduction of $MnO_4^{-}$ to $Mn^{2+}$ and the oxidation of $A^{x+}$ to $AO_3^{-}$.For $MnO_4^{-} ightarrow Mn^{2+}$,the change in oxidation state of $Mn$ is from $+7$ to $+2$,which is a gain of $5$ electrons per mole of $MnO_4^{-}$.For $A^{x+} ightarrow AO_3^{-}$,the oxidation state of $A$ changes from $+x$ to $+5$. The number of electrons lost per mole of $A$ is $(5 - x)$.According to the law of equivalence,the total electrons gained equals the total electrons lost.$1 	imes 5 = 1.25 	imes (5 - x)$$5 = 1.25 	imes (5 - x)$$4 = 5 - x$$x = 5 - 4 = 1$Therefore,the value of $x$ is $1$.

$MnO_4^{-}$ reacts with $A^{x+}$ to form $AO_3^{-}$,$Mn^{2+}$ and $O_2$. One mole of $MnO_4^{-}$ oxidizes $1.25$ moles of $A^{x+}$ to $AO_3^{-}$. What is the value of $x$?

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