## Balancing redox reactions

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

CameronDis2K
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Joined: Wed Feb 20, 2019 12:18 am

### Balancing redox reactions

Hi! I'm still confused on how to decide which element in a multi-element compound you chose that is apparently oxidized or reduced?

Sanjana Munagala_1j
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Joined: Sat Aug 24, 2019 12:17 am
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### Re: Balancing redox reactions

You want to remember that cell potential needs to be positive in order to have a favorable redox reaction. For this reason when you have two redox reactions calculate the E cell using the equation: Ecell=Ecathode - Eanode. Use this equation twice putting the individual cell potentials for the half reactions in place of the either cathode or anode and then switch it. Whichever gives you a (more) positive E cell will be favored and thus the Ecathode half reaction would indicate a reduction reaction and the Eanode half reaction would indicate a oxidation reaction.

Hope that helps!

Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

### Re: Balancing redox reactions

Specifically when you're balancing chemical equations, it's goo to work with an example. Say you have Al(s) + MnCl2(aq) -> AlCl3(aq) + Mn(s). One of the things to recognize is that we really have manganese and aluminum cations here when they dissociate in aqueous solution into their counterparts of Mn2+ and Al3+. So, we can rewrite our chemical equation like this: Al(s) + Mn2+(aq) -> Al3+(aq) + Mn(s)
We can break this equation into two smaller reduction and oxidation equations:
Al(s) -> Al3+(aq) + 3e-
2e- + Mn2+(aq) -> Mn(s)
balancing these chemical equations, we get
2Al(s) -> 2Al3+(aq) + 6e-
6e- + 3Mn2+(aq) -> 3Mn(s)
and thus
2Al(s) + 3Mn2+(aq) -> 2Al3+(aq) + 3Mn(s).
We know from the broken down reduction and oxidation equations that Mn is being reduced from Mn2+ because it is gainnig electrons, whereas Al is being oxidized from a solid state to an aqueous one as it loses electrons.