Question

Balance the following reaction by ion-electron method: $K^{+}MnO_{4}^{-}+H^{+}Cl^{-} \longrightarrow K^{+}Cl^{-}+Mn^{2+}(Cl^{-})_{2}+H_{2}O+Cl_{2}$

Answer 1

First write the reaction skeleton

$HCl\quad +\quad KMn{ O }_{ 4 }\quad \rightarrow \quad KCl\quad +\quad Mn{ Cl }_{ 2 }\quad +\quad { H }_{ 2 }O\quad +\quad { Cl }_{ 2 }$

a) Assign oxidation numbers  

${ H }^{ +1 }{ Cl }^{ -1 }\quad +\quad { K }^{ +1 }{ Mn }^{ +7 }{ O }_{ 4 }^{ -2 }\quad \rightarrow \quad { K }^{ +1 }{ Cl }^{ -1 }\quad +\quad { Mn }^{ +2 }{ Cl }_{ 2 }^{ -1 }\quad +\quad { H }_{ 2 }^{ +1 }{ O }^{ -2 }\quad +\quad { Cl }_{ 2 }^{ 0 }$

b) Identify redox couples

$O: { H }^{ +1 }{ Cl }^{ -1 }\quad \rightarrow \quad { Cl }_{ 2 }^{ 0 }$

$R: { K }^{ +1 }{ Mn }^{ +7 }{ O }_{ 4 }^{ -2 }\quad \rightarrow \quad { Mn }^{ +2 }\quad +\quad 2{ Cl }_{ 2 }^{ -1 }$

c) Combine these redox couples.

$O: { H }^{ +1 }{ Cl }^{ -1 }\quad \rightarrow \quad { Cl }_{ 2 }^{ 0 }$

$R: { K }^{ +1 }{ Mn }^{ +7 }{ O }_{ 4 }^{ -2 }\quad \rightarrow \quad { Mn }^{ +2 }{ Cl }_{ 2 }^{ -1 }$

Step 3. Balance the atoms in each half reaction.  

$O: 2HCl\quad \rightarrow \quad { Cl }_{ 2 }$

$R: KMn{ O }_{ 4 }\quad +\quad 3HCl\quad \rightarrow \quad Mn{ Cl }_{ 2 }\quad +\quad KCl$

b) Balance the oxygen atoms.  

$O: 2HCl\quad \rightarrow \quad { Cl }_{ 2 }$

$R: KMn{ O }_{ 4 }\quad +\quad 3HCl\quad \rightarrow \quad Mn{ Cl }_{ 2 }\quad +\quad KCl\quad +\quad 4{ H }_{ 2 } O$

c) Balance the hydrogen atoms.  

$O: 2HCl\quad \rightarrow \quad { Cl }_{ 2 }\quad +\quad 2 { H }^{ + }$

$R: KMn{ O }_{ 4 }\quad +\quad \quad +\quad 5 { H }^{ + }\quad \rightarrow \quad Mn { Cl }_{ 2 }\quad +\quad KCl\quad +\quad 4{ H }_{ 2 } O$

Step 4. Balance the charge.  

$O: 2HCl\quad \rightarrow \quad { Cl }_{ 2 }\quad +\quad 2 { H }^{ + }\quad +\quad 2{ e }^{ - }$

$R: KMn{ O }_{ 4 }\quad +\quad 3HCl\quad +\quad 5 { H }^{ + }\quad +\quad 5{ e }^{ - }\quad \rightarrow \quad Mn{ Cl }_{ 2 }\quad +\quad KCl\quad +\quad 4{ H }_{ 2 } O$

Step 5. Make electron gain equivalent to electron lost.  

$O: 2HCl\quad \rightarrow \quad { Cl }_{ 2 }\quad +\quad 2 { H }^{ + }\quad +\quad 2{ e }^{ - }$

$R: KMn { O }_{ 4 }\quad +\quad 3HCl\quad +\quad 5 { H }^{ + }\quad +\quad 5{ e }^{ - }\quad \rightarrow \quad Mn { Cl }_{ 2 }\quad +\quad KCl\quad +\quad 4{ H }_{ 2 } O$

$O: 10HCl\quad \rightarrow \quad 5 { Cl }_{ 2 }\quad +\quad 10 { H }^{ + }\quad +\quad 10{ e }^{ - }$

$R: 2KMn { O }_{ 4 }\quad +\quad 6HCl\quad + \quad 10 { H }^{ + }\quad +\quad 10{ e }^{ - }\quad \rightarrow  \quad 2Mn { Cl }_{ 2 }\quad +\quad 2KCl\quad +\quad 8{ H }_{ 2 } O$

Step 6. Add the half-reactions  

$2KMn { O }_{ 4 }\quad +\quad 16HCl\quad +\quad 10 { H }^{ + }\quad +\quad 10{ e }^{ - } \quad \rightarrow \quad 5 { Cl }_{ 2 }\quad + \quad 2Mn { Cl }_{ 2 } \quad +\quad 10 { H }^{ + }\quad + \quad 2KCl\quad +\quad 10{ e }^{ - } \quad + \quad 8{ H }_{ 2 } O$

Step 7. Simplify the equation.  

$2KMn { O }_{ 4 }\quad +\quad 16HCl\quad \rightarrow  \quad 5 { Cl }_{ 2 }\quad +\quad 2Mn { Cl }_{ 2 } \quad +\quad 2KCl\quad +\quad 8{ H }_{ 2 } O$

Step 8. Charge on L.H.S is nullified by charge on R.H.S