# How to use the Nernst Equation

So I understand the Nernst Equation, but I don't have some info. (Please bear with me, I'm in secondary.)

I'm doing a Cu - Al cell. My E cell is negative (-1.33). Does this have an affect on the equation?

What exactly does E0 stand for? I understand how to find E cell, but do I square it by something? I can't find online how many electrons are swapped. How many electrons are swapped?

How do you factor in the pH? I've calculated the pH and the number of ions in it, but where do i add this in on the equation? Please lay it out how you would in the equation so I can see. (I have numerous pH's)

I'm stuck on R (gas constant, which is 8.31 (volt-coulomb)/(mol-K)). I can't figure out the volt-coulomb as i don't know the voltage. How can I calculate the coulomb? Then the volt-coulomb? And is Mol-K (Mole)(Kelvin)? If so, do I factor in the temperature in Kelvins twice?

For N (n = number of moles of electrons exchanged in the electrochemical reaction) do i just multiply the moles per electron? Finally - What exactly am I dividing to calculate Q? (How do I calculate Q?)

Could you do one full equation as a template for me please? One pH level is 4.3 Please explain this all in basic jargon - or just normal English if you prefer. My vocabulary is limited. Thank you!

• I can help more if you have an example of a question from past-year competition archives. – orthocresol Sep 26 '15 at 13:30
• I have everything sorted except N- Could you please directly answer how many electrons are exchanged- I'm unsure if it's 6, 3, or 5 (adding them both together). Thank you! – Kiwi Sep 26 '15 at 14:00

Asking the faceless masses of the internet for help that you need very quickly can be a dangerous proposition, but let's see where this goes.

Here's the Nernst Equation: $$E=E^\circ - \dfrac{RT}{nF}\ln{Q}$$

• $E$ is the measured cell potential
• $E^\circ$ is the standard cell potential, assuming 1.0 molar concentration, 1 atmosphere of pressure, and 25 degrees Celsius
• $R$ is the ideal gas constant - $8.314\ \mathrm{\frac{V\cdot C}{K\cdot mol}}$
• $T$ is the temperature
• $n$ is the number of electrons transferred
• $F$ is the Faraday constant $9.648\times 10^4\ \mathrm{\frac{C}{mol}}$
• $\ln$ is the natural logarithm
• $Q$ is the reaction quotient, which is related to the amount of reactants in solution.

Finding $E^\circ$

You look this value up. Actually you look up the values of $E^\circ$ for your half-reactions.

You say you are doing copper-aluminum, so your half-reactions are probably:

$$\ce{Cu^2+ + 2e- -> Cu}$$ $$\ce{Al^3+ + 3e- -> Al}$$

Here is a link to Wikipedia's Standard Electrode Potential Data Page

$$\begin{array}{r|c} \mathrm{half-reaction} & E^\circ/\mathrm{V}\\ \hline \ce{Cu^2+ + 2e- -> Cu} & +0.337 \\ \ce{Al^3+ + 3e- -> Al} & -1.662\\ \end{array}$$

How many electrons are exchanged?

You determine this number from your half reactions.

$$\begin{array}{r|c|l} \mathrm{half-reaction} & E^\circ/\mathrm{V} & n\\ \hline \ce{Cu^2+ + 2e- -> Cu} & +0.337 & 2 \\ \ce{Al^3+ + 3e- -> Al} & -1.662 & 3\\ \end{array}$$

How do you factor in pH?

You only factor in pH if $\ce{H+}$ or $\ce{OH-}$ appear in your reactions. In this case, neither does. However, high pH might shut down your reaction since both $\ce{Al(OH)3}$ and $\ce{Cu(OH)2}$ are low solubility.

How do I factor in the number of ions?

I suppose you mean the concentrations of $\ce{Cu^2+}$ and $\ce{Al^3+}$? These concentrations impact $Q$, the reaction quotient. For simple half reactions like yours, $Q=\frac{1}{\mathrm{ion\ concentration}}$

Units

The volt-coulomb and kelvin-mole units on $R$ are just what they are, units. You do not need to determine them; they are part of the value of $R$. These units are necessary to that all of the other units cancel leaving just volts as the unit of $\frac{RT}{nF}$

Putting it all together

You now have what you need to determine the half-reaction potentials at your nonstandard values. Just plug in what you have for $E^\circ$, $R$, $T$, $n$, $F$, and $Q$. Then you can calculate your cell potential.

Positive cell potential is spontaneous, so we want to plug in values for reduction (as written) and oxidation (reversed) to get the largest possible positive number. In this case (in which I use the standard potentials since i do not know your concentrations or temperatures):

$$E^\circ =\mathrm{ +0.337\ V -(-1.662\ V) = +1.999\ V}$$

Copper is reduced; aluminum is oxidized.

However, likely you are doing this experiment to calculate or verify something, likely either $E^\circ$ or ion concentration. If this is the case, leave that quantity as a variable and solve for it.

However, your $E$ is negative, which suggests electrolysis. Is this what you are doing?

• Actually the value of $n$ in this case depends on how you combine the half-equations - so for the reaction $\ce{3 Cu + 2 Al^{3+} -> 3 Cu^{2+} + 2 Al}$ (or the reverse) you'd have $n = 6$, but for the reaction $\ce{3/2 Cu + Al^{3+} -> 3/2 Cu^{2+} + Al}$ you'd have $n = 3$. In the end the maths works out to be the same because of how the reaction quotient $Q$ is defined – orthocresol Sep 26 '15 at 13:29
• I have everythng sorted except N- Could you please directly answer how many electrons are exchanged- I'm unsure if it's 6, 3, or 5 (adding them both together). Thank you! – Kiwi Sep 26 '15 at 13:58
• @Kiwi - see orhocresol's comment. My explanation focuses on keeping the half-reactions separate for as long as possible. Then you do not need to worry about the number of electrons exchanged in the full reaction. – Ben Norris Sep 26 '15 at 14:24
• @Kiwi - refer to my comment above. In the first equation I wrote you have 3 copper atoms giving up 2 electrons each for a total of 6. These 6 electrons are accepted by 2 aluminium ions. Therefore $n = 6$. Note that $n$ is a stoichiometric coefficient, and although it's how I described it above, it's not a "number of electrons transferred". It's more like "number of moles of electrons transferred per mole of the reaction as written". By that I mean, the proper explanation should be: if you react 3 moles of Cu and 2 moles of Al$^{3+}$, 6 moles of electrons will be transferred, and hence $n = 6$. – orthocresol Sep 26 '15 at 15:24
• @Kiwi the reason for the distinction is because obviously you can't have 1.5 copper atoms so the same explanation fails for the second equation I wrote.... but if you have 1.5 moles of copper then 3 moles of electrons are transferred. Ben's method of keeping the half-equations separate is also entirely valid and personally I think it is a more "proper" explanation as $n$ is an explicit stoichiometric coefficient in a chemical equation. However you do have to be able to keep track of which half-reaction has which value of $n$ as usually they will not be the same (as it is in this case). – orthocresol Sep 26 '15 at 15:29