1.

Balance the following equation by ion-electron method.

Answer»

Solution :`S_(2)O_(3)^(2-)+I_(2)rarrS_(2)O_(4)^(2-)+SO_(2)+I^(-)`
Oxidation half REACTION:
`underset(+2)(S_(2)O_(3)) RARR underset(+3)(S_(2)O_(4)^(2-)) + cancel(2e^(-))` ................(1)
Reduction half reaction:
`I_(2)+cancel(2e^(-))rarr2I^(-)` ................(2)
Add equation (1) + (2)
`S_(2)O_(3)^(2-) + I_(2) rarr S_(2)O_(4)^(2-)+2I^(-)`
To balance, `SO_(2)` is added on RHS of the equation.
`S_(2)O_(3)^(2-)+I_(2)rarrS_(2)O_(4)^(2-)`
To balance oxygen atom, `S_(2)O_(3)^(2-)` and `SO_(2)` is MULTIPLIED by 2.
`2S_(2)O_(3)^(2-) + I_(2) rarr S_(2)O_(4)^(2-)+2I^(-)+2SO_(2)`
(ii) `Sb^(3+) + MnO_(4)^(-) rarr Sb^(5+) + Mn^(2+)`
Oxidation half reaction:
`Sb^(3+)rarr Sb^(5+) + 2e^(-)` .................(1)
Reduction half reaction:
`underset((+7))(MnO_(4)^(-))+ 5e^(-)rarr Mn^(2+)` .................(2)
In equation (2),` H_(2)O` is added on LHS to balance oxygen atom.
`MnO_(4)^(-) +5e^(-) rarr Mn^(2+) + 4H_(2)O` .................(3)
To balance Hydrogen atoms, `H^(+)` is added on RHS.
`MnO_(4)- + 5e^(-) + 8H^(+) rarr Mn^(2+) + 4H_(2)0` .................(4) Equation (4) is multiplied by 2 and equation (1) is multiplied by 5 to equalise the electrons gained and electrons lost.
(1) `=gt 5Sb^(3+) rarr 5Sb^(5+)+ 10e^(-)` .................(5)
(4)`=gt 2MnO_(4)^(-)+ 10e^(-) + 16H^(+) rarr 2Mn^(2+) + 8H_(2)O` ...........(6)Add (5) and (6)
`5Sb^(3+) +2MnO_(4)^(-) "+ 16H^(+) rarr 5Sb^(5+)+ 2Mn^(2+)+ 8H_(2)O`
(iii) `MnO_(4)^(-) + I^(-) rarr MnO_(2) + I_(2)`
Oxidation half reaction:
`2I^(-)+2e^(-)rarrI_(2)` ................ (1)
Reduction half reaction:
`MnO_(4)^(-) rarr MnO_(2) + 3e^(-)` ................(2)
Equation (1) is multiplied by 3. `6I^(-) + cancel(6e^(-)) rarr 3I_(2)` ...........(3)
Equation (2) is multiplied by 2.
`2MnO_(4)^(-) rarr 2MnO_(2) + cancel(6e)` .....................(4)
`2MnO_(4)^(-)+6I^(-) rarr 2MnO_(2) + 3I_(2)`
To balance oxygen and hydrogen atoms, `H^(+)` RHS and `H_(2)O `is added on LHS.
`2MnO_(4)^(-)+6I^(-)+4H^(+)rarr2MnO_(2)+2I_(2)+2H_(2)O`
I acidic medium
(iv) `MnO_(4)^(1)+Fe^(2+)rarrMn^(2+)+Fe^(3+)`
Oxidation half reaction:
`Fe^(2+)RARRFE^(3+)+e^(-)`.....................(1)
Reduction half reaction:
`underset((+7))(MnO_(4)^(-))+5e^(-)rarrMn^(2+)`...........(2)
`(1)xx5 =gt 5Fe^(2+)rarr5Fe^(3+)+cancel(5e^(-))`
(2)`=gtMnO_(4)^(-)+ cancel(5e^(-))rarrMn^(2+)`.......Add equation (1) and (2)
`MnO_(4)^(-)+5Fe^(2+)rarrMn^(2+)+5Fe^(3+)`
To balance oxygen, `H^(+)` is added on RHS and `H_(2)O` is added on LHS
`MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarrMn^(2+)+5Fe^(3+)+4H_(2)O`
(v) `Cr(OH)_(4)^(-) + H_(2)O_(2) rarr CrO_(4)^(2-)`
Oxidation half reaction:
`underset((+3))(Cr(OH)_(4)^(-)) rarr underset((+1))(CrO_(4)^(2-)) + 3e^(-)` ...........................(1)
Reduction half reaction:
`underset((+2))(H_(2)O_(2)) + e^(-) rarr underset((+1))(H_(2)O)`..............(2)
`(2)xx3=gt3H_(2)O_(2)+cancel(3e^(-))rarr3H_(2)O`
(1)`=gtCr(OH)_(4)^(-)rarrCrO_(4)^(2-)+3e^(-)`Adding Equation (1) and(2)
`Cr(OH)_(4)^(-) + 3H_(2)O_(2)rarrCrO_(4)^(2-)+3H_(2)O`
To balance oxygen and hydrogen atoms, `OH^(-)` and `H_(2)O` are added
`2Cr(OH)_(4)^(-) + 3H_(2)O_(2)+2OH^(-)rarr2CrO_(2)^(2-)+8H_(2)O`


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