Question

Write a balanced half-reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution. Be sure to add physical state symbols where appropriate.

Answer 1

Answer:
`MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^(2+)(aq)+2H_2O(l)`

Step-by-step explanation:

Reduction is a process where electrons are gained and acidic solution means presence of
`H^+` ions.

Reduction of
`MnO_2` to
`Mn^(2+)`

Mn is in +4 oxidation state in
`MnO_2` which goes to +2 state in
`Mn^(2+)` by gain of 2 electrons.

`MnO_2(s)\rightarrow Mn^(2+)(aq)`

In order to balance oxygen atoms:

`MnO_2(s)\rightarrow Mn^(2+)(aq)+2H_2O`

In order to balance hydrogen atoms:

`MnO_2(s)+4H^+(aq)\rightarrow Mn^(2+)(aq)+2H_2O(l)`

In order to balance charges:

`MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^(2+)(aq)+2H_2O(l`

Thus the net balanced half reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:

`MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^(2+)(aq)+2H_2O(l)`