In the redox reaction xMnO + yPbO_2 + zHNO_3 | vedclass

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Question

(A) The given reaction is: $xMnO + yPbO_2 + zHNO_3 	o HMnO_4 + Pb(NO_3)_2 + H_2O$. Assign oxidation states: $Mn$ in $MnO$ is $+2$,in $HMnO_4$ is $+7$

Vedclass · Accepted Answer

$x = 2, y = 5, z = 10$

Vedclass · Answer

(A) The given reaction is: $xMnO + yPbO_2 + zHNO_3 	o HMnO_4 + Pb(NO_3)_2 + H_2O$. Assign oxidation states: $Mn$ in $MnO$ is $+2$,in $HMnO_4$ is $+7$. Change in $Mn = +5$. $Pb$ in $PbO_2$ is $+4$,in $Pb(NO_3)_2$ is $+2$. Change in $Pb = -2$. To balance the electron transfer,multiply $Mn$ by $2$ and $Pb$ by $5$: $2MnO + 5PbO_2 + zHNO_3 	o 2HMnO_4 + 5Pb(NO_3)_2 + H_2O$. Now balance $N$ atoms: There are $5 	imes 2 = 10$ nitrogen atoms on the right,so $z = 10$. Finally,balance $H$ and $O$: $2MnO + 5PbO_2 + 10HNO_3 	o 2HMnO_4 + 5Pb(NO_3)_2 + 4H_2O$. Thus,$x = 2, y = 5, z = 10$.

In the redox reaction $xMnO + yPbO_2 + zHNO_3 \to HMnO_4 + Pb(NO_3)_2 + H_2O$,determine the values of $x, y,$ and $z$.

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