Respuesta :
Answer:
The balanced reaction is:
[tex]8MnO_{4}^{-}+14H^{+}+5H_{2}S--->8Mn^{+2}+12H_{2}O+5SO_{4}^{-2}[/tex]
The coefficient is "5".
Explanation:
The reaction is in acidic medium.
The oxidized species is S and the reduced one is Mn
The reduction half equation is:
[tex]MnO_{4}^{-}--->Mn^{+2}[/tex]
The Mn atom is already balanced, let us balance oxygen atom by adding four water molecules on right side and then hydrogen atom by adding hydrogen ions on other side.
[tex]MnO_{4}^{-}+8H^{+}--->Mn^{+2}+4H_{2}O[/tex]
The charge will be balanced by adding electrons to the side with more positive charge.
[tex]5e+MnO_{4}^{-}+8H^{+}--->Mn^{+2}+4H_{2}O[/tex]
The oxidation half reaction is:
[tex]H_{2}S--->SO_{4}^{-2}[/tex]
Again we will balance oxygen with water and hydrogen with hydrogen ions and charge by electrons
[tex]H_{2}S+4H_{2}O--->SO_{4}^{-2}+10H^{+}+8e[/tex]
Now we will multiply the oxidation half reaction with "5" and reduction half reaction with "8" (to balance the electrons) and will add them.
The resulting equation will be:
[tex]8MnO_{4}^{-}+14H^{+}+5H_{2}S--->8Mn^{+2}+12H_{2}O+5SO_{4}^{-2}[/tex]
