## balancing redox reactions where the oxidizing agent and reducing agent are the same

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

### balancing redox reactions where the oxidizing agent and reducing agent are the same

I'm having trouble balancing redox reactions where the oxidizing agent and reducing agent are the same. Is there a different process for these types of reactions?
For example, Br2(l)→BrO3−(aq)+Br−(aq) has Br2 as both reducing and oxidizing agents. Also P4(s)→H2PO2−(aq)+PH3(aq) has P4 as both reducing and oxidizing agents.

Sebastian Lee 1L
Posts: 157
Joined: Fri Aug 09, 2019 12:15 am
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### Re: balancing redox reactions where the oxidizing agent and reducing agent are the same

In these cases, you would have two separate half-reactions with the same reactant for both the oxidation half-reaction and reduction half-reaction.

In the case of Br2 that you gave, the oxidation half-reaction would be $Br_{2} (l)+12OH^{-}(aq)\rightarrow 2BrO_{3}^{-}(aq)+6H_{2}O(l)+10e^{-}$. You start with Br2 -> BrO3- and then add water to the left hand side to balance oxygens, and then water on the right with hydroxide on the left to balance hydrogens (more detailed explanation in the textbook). Finally you add the number of electrons released.

The reduction half-reaction would be $Br_{2}(l)+2e^{-}\rightarrow 2Br^{-}(aq)$. Again, the reactant is the same as the oxidation reaction but this time it is being reduced to an anion.

When you combine the reactions (multiplying by needed coefficients to cancel out electrons), you should find the correct balanced reaction. Notice that some of the Br2 becomes oxidized (1 mole) while some of the Br2 becomes reduced (5 moles).

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

### Re: balancing redox reactions where the oxidizing agent and reducing agent are the same

Sebastian Lee 1L wrote:In these cases, you would have two separate half-reactions with the same reactant for both the oxidation half-reaction and reduction half-reaction.

In the case of Br2 that you gave, the oxidation half-reaction would be $Br_{2} (l)+12OH^{-}(aq)\rightarrow 2BrO_{3}^{-}(aq)+6H_{2}O(l)+10e^{-}$. You start with Br2 -> BrO3- and then add water to the left hand side to balance oxygens, and then water on the right with hydroxide on the left to balance hydrogens (more detailed explanation in the textbook). Finally you add the number of electrons released.

The reduction half-reaction would be $Br_{2}(l)+2e^{-}\rightarrow 2Br^{-}(aq)$. Again, the reactant is the same as the oxidation reaction but this time it is being reduced to an anion.

When you combine the reactions (multiplying by needed coefficients to cancel out electrons), you should find the correct balanced reaction. Notice that some of the Br2 becomes oxidized (1 mole) while some of the Br2 becomes reduced (5 moles).

I did this for question 6K.5b and I didn't get the same answer as in the textbook which makes me confused

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