## Finding the n value for a redox reaction!

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Regina Chi 2K
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Joined: Fri Sep 26, 2014 2:02 pm

### Finding the n value for a redox reaction!

For problem 4 in the 2013 practice final exam, I was wondering how they determined the n value to be 4? I understand that we can get the n value by looking at the redox reaction, but looking at the equation, I would think that it is 2, and not 4. I was thinking that it might be coming from the half-reactions from the formula sheet, but once again, how would we know which one to use (because there is a lot of stuff with H2O and OH-)...

Neil DSilva 1L
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Joined: Fri Sep 26, 2014 2:02 pm

### Re: Finding the n value for a redox reaction!

The overall redox reaction given in the problem is $2Fe(s) + 2H_{2}O(l) + O_{2}(aq) \rightarrow 2Fe(OH)_{2}(s)$.

The way that I was able to determine which half reaction to use was by looking at the products and reactants given and the overall standard cell potential given.

If you break up the overall reaction into two half reactions (it may help to separate all the reactants and products into ions), you would have:
$2H_{2}O + O_{2} + 4e^{-} \rightarrow 4OH^{-}$ (note: this is appears on the data sheet with standard cell potential 0.40 V)
and $2Fe + \rightarrow 2Fe^{2+} + 4e^{-}$ (note: this appears on the data sheet, flipped and halved with standard cell potential -.044 V)

You know these two half reactions are correct, because the standard cell potential determined from the data sheet is the same as the one given in the problem.

Don't forget to balance your half reactions. We need to multiply the Fe half reaction by two so that it is balanced and appears the same way as in the problem. This gives us an overall n value of 4.

Chem_Mod
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### Re: Finding the n value for a redox reaction!

Well done Neil!

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