Question

Write half-reactions for the oxidation and reduction process for each of the following.

a. Fe2+ + MnO4 - Fe3+ + Mn2+
b. Sn2+ + IO3 - Sn4+ + I-
c. S2- + NO3 - S + NO
d. NH3 + NO2 N2 + H2O

Answer 1

a)
`Fe^(2+)`⇒
`Fe^(3+) + e^(-)`

`MnO_(4) ^(-)` + 2
`H_(2)O` + 3
`e^(-)`⇒
`MnO_(2) + 4OH^(-)`

b)
`Sn^(2+)` ⇒
`Sn^(4+) + 2e^(-)`

`IO_(3) ^(-)  + 6H^(+) + 5e^(-)`⇒
`I_(2) + 3H_(2)O`

`I_(2) + 2e^(-)`⇒
`2I^(-)`

c)
`S^(2-)`⇒
`S+ 2e^(-)`

`NO_(3) ^(-) + 4H^(+) + 3e^(-)`⇒
`NO + 2H_(2)O`

d)
`2NH_(3)  + 6OH^(-)`⇒
`N_(2) + 6H_(2)O +6e^(-)`

`2NO_(2)  + 8H^(+) + 8e^(-)`⇒
`N_(2) + 4H_(2)O`

Step-by-step explanation:

a)
`Fe^(2+)`⇒
`Fe^(3+) + e^(-)`

`MnO_(4) ^(-)` + 2
`H_(2)O` + 3
`e^(-)`⇒
`MnO_(2) + 4OH^(-)` half reaction suggest that you need water to reaction occurs.

b)
`Sn^(2+)` ⇒
`Sn^(4+) + 2e^(-)`

`IO_(3) ^(-)  + 6H^(+) + 5e^(-)`⇒
`I_(2) + 3H_(2)O`

`I_(2) + 2e^(-)`⇒
`2I^(-)` half reaction suggest that you need acid to reaction acurrs.

c)
`S^(2-)`⇒
`S+ 2e^(-)`

`NO_(3) ^(-) + 4H^(+) + 3e^(-)`⇒
`NO + 2H_(2)O`

d)
`2NH_(3)  + 6OH^(-)`⇒
`N_(2) + 6H_(2)O +6e^(-)`

`2NO_(2)  + 8H^(+) + 8e^(-)`⇒
`N_(2) + 4H_(2)O` note that you can obtain the total reaction by miltiplying first half reaction by 4 and second half reaction by 3.