Question

In the redox reaction: $xMnO+yPb{O}_2+zHN{O}_3\to HMn{O}_4+Pb(N{O}_3{)}_2+H_{2}O$.

• A. $x=2,y=5,z=10$
• B. $x=2,y=7,z=8$
• C. $x=2,y=5,z=8$
• D. $x=2,y=5,z=5$

Answer 1

Answer: (A)

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

Write the reaction in two half reactions,oxidation half and reduction half as:

Oxidation:$MnO\to Mn{O}_4^{-}$

Reduction:$Pb{O}_2\to P{b}^{+2}$

Balance all atoms other than $O$ and $H$.

$MnO\to Mn{O}_4^{-}$

$Pb{O}_2\to P{b}^{+2}$

Now balance the oxygen atoms by adding $H_{2}O$ molecules.

$MnO+3H_{2}O\to Mn{O}_4^{-}$

$Pb{O}_2\to P{b}^{+2}+2H_{2}O$

Now balance hydrogen atoms by adding $H^{+}$ ions.

$MnO+3H_{2}O\to Mn{O}_4^{-}+6H^{+}$

$Pb{O}_2+4H^{+}\to P{b}^{+2}+2H_{2}O$

To balance the charge,add electrons to more positive side to equal the less positive side of the half reaction.

$MnO+3H_{2}O\to Mn{O}_4^{-}+6H^{+}+5e^{-}$

$Pb{O}_2+4H^{+}+2e^{-}\to P{b}^{+2}+2H_{2}O$

Now multiply oxidation half reaction by 2 and reduction half reaction by 5.Add both the reactions we will get

$2MnO+5Pb{O}_2+8H^{+}\to 2Mn{O}_4^{-}+5P{b}^{2+}+4H_{2}O$

To change the ionic equation to original one we should add hydrogen ions equal to negative charge and nitrate ions equal to positive charge.

$2MnO+5Pb{O}_2+8HN{O}_3\to 2HMn{O}_4+5P{b}_{(}N{O}_3{)}_2+4H_{2}O$