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We've been studying redox reactions. So far, everything has been under standard conditions. We started by balancing redox reactions. We had the half-reaction method and the oxidation number change method. Then, we moved into balancing redox in acidic and basic conditions. But, we always assumed standard one molar concentrations. Finally, we started to calculate voltages for voltaic cells. But even there, we used the standard electrode potentials. Are batteries ever under non-standard conditions?

Let's consider a rechargeable battery-powered device. Maybe think of your phone. What's happening as you use it? Electrons flow from the anode to the cathode. We use the work generated by the current to power the phone's functions. But what's happening inside the battery? The anode is oxidized. So, it transitions from a solid to aqueous. The cathode gets reduced. Some of its ions gain electrons. They go from being ions to the solid phase. When your phone is half-charged, do you have one molar concentration? Probably not. What happens when your phone reaches twenty percent charged? It enters low power mode. It minimizes functionality to conserve energy. So, we. might expect a reduced voltage. Do we have an equation for that?

The Nernst Equation

Let's start with the Gibbs free energy equation for non-standard conditions. Do you remember the connection between the free energy and the equilibrium constant? We learned it way back in thermodynamics. Remember, we are not at standard conditions. So, we use the reaction quotient, (Q)(Q), instead of the equilibrium constant (KeqK_{eq}).

ΔG=ΔGo+RTln⁡(Q)\Delta G= \Delta G^{o} + RT\ln{(Q)}

Previously, we connected the voltage and free energy change. So, we can replace both ΔG\Delta G values with −nFEcell-nFE_{cell}.

−nFE=−nFEo+RTln⁡(Q)-nFE= -nFE^{o} + RT\ln{(Q)}

We want to isolate EE. So, we can divide both sides of the equation by −nF-nF.

E=Eo−RTnFln⁡(Q)E=E^{o} - \frac{RT}{nF}\ln{(Q)}

Typically, we write the Nernst equation as a base-10 logarithm. So, we can use ln⁡(x)=2.303log⁡(x)\ln{(x)}=2.303\log{(x)}, to simplify.

E=Eo−2.303RTnFlog⁡(Q)E=E^{o} - \frac{2.303RT}{nF}\log{(Q)}

When the temperature is 298.15K, we can simplify the equation even more.

E=Eo−0.0592nlog⁡(Q) when T=298.15KE=E^{o} - \frac{0.0592}{n}\log{(Q)}\, when\, T=298.15K

Example Problem

In acidic conditions, copper, a transition metal, reacts with nitrate ions. The products are copper ions and nitrogen monoxide. What's the cell's voltage at 298.15K298.15K when [NO32−]=2.0M, [H+]=1.0M, [Cu2+]=.01M, PNO=1.0atm[NO_{3}^{2-}]=2.0M,\, [H^{+}]=1.0M,\, [Cu^{2+}]=.01M, \,P_{NO}=1.0atm?

To start, we should write the balanced chemical equation. Nitrate reduction always requires an acid. So, most standard electrode potential tables give you the acid half-reaction. But, if you didn't have it, you could use the steps in balancing redox reactions in acidic solutions.

3Cu+2NO3−+8H+→3Cu2++2NO+4H2O3Cu + 2NO_{3}^{-} +8 H^{+} \rightarrow 3Cu^{2+} + 2NO + 4H_{2}O

We can split the reaction into two half-reactions.

3Cu2++6e−→3Cu(s) Eo=0.34V3Cu^{2+}+ 6e^{-} \rightarrow 3Cu_{(s)}\, E^{o}=0.34V

2NO3−+8H++6e−→2NO(g)+4H2OEo=0.96V2NO_{3}^{-}+ 8H^{+} + 6e^{-}\rightarrow 2NO_{(g)} + 4H_{2}O E^{o}=0.96V

Now we can calculate the standard potential for this cell.

Ecello=Ecathodeo−Eanodeo=0.96V−0.34V=0.62VE_{cell}^{o}=E_{cathode}^{o} - E_{anode}^{o}=0.96V-0.34V=0.62V

Now we can write the Nernst equation. Because we are at 298.15K298.15K, we can use the shortened expression.

E=Eo−0.0592nlog⁡(Q)=0.62V−0.05926log⁡([Cu2+]3[NO3−]2[H+]8)E=E^{o} - \frac{0.0592}{n}\log{(Q)} =0.62V- \frac{0.0592}{6}\log{(\frac{[Cu^{2+}]^{3}}{[NO_{3}^{-}]^{2}[H^{+}]^{8}})}

=0.62V−0.00987log⁡(.00025)=0.62V−(−.036V)=0.66V.=0.62V-0.00987\log{(.00025)}=0.62V-(-.036V)=0.66V.

In this problem, we didn't have 1.0 molar concentrations. We weren't at standard conditions! What was the result? Our voltage was larger than the standard state voltage!

A Little History of Walther Nernst

Walther Nernst is an unbelievably famous physical chemist. In 1920 he won the Nobel prize for his work in thermodynamics. What was his contribution? He only developed the Third Law of Thermodynamics! Do you remember it? It states, "the entropy of a pure crystal lattice, at absolute zero, equals zero." Why is it so important? It was foundational for determining a chemical reaction's free energy and equilibrium constant. As we just learned, he wasn't a one-hit-wonder. He applied these concepts to electrochemistry. Now we know the Nernst equation! A powerful tool for voltaic cells under non-standard conditions!

Video

(Video) 19.5 How to Calculate Nonstandard Cell Potential [Nernst Equation] | General Chemistry

by Chad's Prep

In this video, Chad explains the Nernst Equation. It allows you to calculate the electrode potential at nonstandard conditions. First, he explains the equation. Then, he shows you an example calculation. Finally, he uses the example calculation to explain how we can use the equation qualitatively. This is an excellent explanation of the Nernst Equation.

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