## Balancing redox reactions

Anaranjo
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### Balancing redox reactions

Balance the following skeletal equations by using oxidation and reduction half-reactions. Identify the oxidizing agent and reducing agent in each reaction.

Mn^2+(aq)+Br2(l)->MnO4^-(aq)+Br(aq)

Chem_Mod
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### Re: Balancing redox reactions

From page 564-565 of textbook, 6th ed.

1. Write the skeletal equation for both parts:
$MnO_{4}^{-}_{(aq)}\rightarrow Mn^{2+}_{(aq)}$
$Br_{2}_{(l)}\rightarrow Br_{(aq)}$
2. Balance all elements except H and O
$MnO_{4}^{-}_{(aq)}\rightarrow Mn^{2+}_{(aq)}$
3. Balance the O atoms by adding $H_{2}O$
$MnO_{4}^{-}_{(aq)}\rightarrow Mn^{2+}_{(aq)}+4H_{2}O$
4. Balance the H atoms by adding $H^{+}$
$MnO_{4}^{-}_{(aq)}+8H^{+}\rightarrow Mn^{2+}_{(aq)}+4H_{2}O_{(aq)}$
5. Balance the net charges by adding electrons.
$MnO_{4}^{-}_{(aq)}+8H^{+}+5e^{-}\rightarrow Mn^{2+}_{(aq)}+4H_{2}O$

Repeat process for Bromine. Figure out which species is gaining electrons (reduction) which species is losing electrons (oxidation). In this case Manganese is being reduced (+7 going to +2 oxidation number)

Sarai_Ferrer_3C
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Joined: Fri Jun 17, 2016 11:28 am

### Re: Balancing redox reactions

Can someone explain to me the reason why we balance reactions under basic conditions differently than when balancing reactions under acidic conditions?

Chem_Mod
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### Re: Balancing redox reactions

Because in acidic conditions there will be protons available, and in basic conditions there are hydroxy anions. There would be no protons in basic conditions because the base would have absorbed this already.