Answer: [tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]
Explanation:
Reduction is a process where electrons are gained and acidic solution means presence of [tex]H^+[/tex] ions.
Reduction of [tex]MnO_2[/tex] to [tex]Mn^{2+}[/tex]
Mn is in +4 oxidation state in [tex]MnO_2[/tex] which goes to +2 state in [tex]Mn^{2+}[/tex] by gain of 2 electrons.
[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)[/tex]
In order to balance oxygen atoms:
[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)+2H_2O[/tex]
In order to balance hydrogen atoms:
[tex]MnO_2(s)+4H^+(aq)\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]
In order to balance charges:
[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l[/tex]
Thus the net balanced half reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:
[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]
