In the previous post, we talked about the Nernst equation which is used to calculate the cell potential under nonstandard conditions based on E^o_cell and the log of the reaction quotient.

where E^o is the cell potential under standard conditions, n is the moles of the electrons in the half-reactions, and Q is the reaction quotient

Another version of the equation is when the natural logarithm of the Q is used:

Keep in mind that both equations are applicable to reactions carried out at 25 ^oC (298 K). 

Notice also that the cell potential under nonstandard conditions differs from the one under standard conditions, only based on the quotient.

Remember, Q has the same expression as the equilibrium constant and the only difference is that we use the given concentrations rather than the ones at the equilibrium:

Reaction quotient, Q tells us whether the forward or reverse reactions are going to be favorable based on the given concentrations of the reactants and products. The driving force here is that the system always tries to reach an equilibrium where Q becomes equal to the K:

Feel free to review the concept of reaction quotient and equilibrium constant by clicking on the highlighted sections, but here is the summary of what we need to know about Q and the direction of the equilibrium:

• Q < K Reaction tends to form more products.
• Q > K Reaction tends to form more reactants.
• Q = K Reaction is already at equilibrium.

Applying this to the Nernst equation, we can write the following correlations between the cell potential and the reaction quotient:

• When Q < 1, it means [reactant] > [product], ln Q < 0, so E_cell > E°_cell.
• When Q = 1, it means [reactant] = [product], ln Q = 0, so E_cell = E°_cell.
• When Q > 1, it means [reactant] < [product], ln Q > 0, so E_cell < E°_cell.
• When Q = K, it means [product] >> [reactant], ln Q = E°_cell, so E_cell = 0 .

Remember, we do not include the liquids and solids, even when they are the electrodes, in the expression for Q as it only contains those species with concentrations (and/or pressures) that can change during the reaction.

For example, in the reaction between manganese and lead ion, the Mn and Pb electrodes do not appear in the expression for Q:

Mn(s) + Pb²⁺(aq) → Mn²⁺(aq) + Pb(s) 

\[E\; = \,{E^o}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[M}}{{\rm{n}}^{{\rm{2 + }}}}{\rm{]}}}}{{{\rm{[P}}{{\rm{b}}^{{\rm{2 + }}}}{\rm{]}}}}\]

Nernst Equation – Example

A voltaic cell runs the following redox reaction:

Sn²⁺(aq) + Mg(s) → Sn(s) + Mg²⁺(aq)

a) For each concentration set, determine, without using the Nernst equation, whether E_cell is larger or smaller than E°_cell and b) Calculate the E_cell:

1) when [Sn²⁺] = 1.0 Mand [Mg²⁺] = 2.5 M

2) when [Sn²⁺] = 3.0 Mand [Mg²⁺] = 1.0 M

1) [Sn²⁺] = 1.0 Mand [Mg²⁺] = 2.5 M

For part (a) 

To estimate if E_cell is larger or smaller than E^o_cell, we need to look at the reaction quotient.

The concentration of the product, Mg²⁺ is higher than that of Sn²⁺ and therefore, according to the Le Chatelier’s principle, the tendency of the forward reaction will be less than at standard conditions since Q > 1.

\[Q\, = \,\frac{{[{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}]}}{{[{\rm{S}}{{\rm{n}}^{{\rm{2 + }}}}]}}\, = \,\frac{{2.5}}{1}\, = \,2.5\]

This also means that E_cell is smaller than E°_cell. Remember the pattern: if Q > 1, E_cell < E°_cell, and if Q < 1, E_cell > E°_cell. 

Part (b). To confirm this assessment, we need to use the Nernst equation and calculate E_cell.

\[E\; = \,{E^o}\; – \,\frac{{0.0257\;{\rm{V}}}}{n}\,\ln \,Q\]

Remember, V is just a unit here, and if it confuses you, you can write the equation as:

\[E\; = \,{E^o}\; – \,\frac{{0.0257}}{n}\,\ln \,Q\]

First, calculate the E°_cell by writing the half-reactions and looking up each cell potential. We have a separate post on calculating the cell potential at standard conditions so feel free to check that for more details.

Sn²⁺(aq) + 2e^– → Sn(s)    E^o = -0.14 V

Mg(s) → Mg²⁺(aq) + 2e^–  E^o = +2.37 V

E^o_cell = -0.14 V + 2.37 V = 2.23 V

And know, we can plug the numbers into the Nernst equation and determine the E_cell:

\[E\; = \,{E^o}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[M}}{{\rm{g}}^{{\rm{2 + }}}}{\rm{]}}}}{{{\rm{[S}}{{\rm{n}}^{{\rm{2 + }}}}{\rm{]}}}}\]

\[E\; = \,2.23\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[2}}{\rm{.5]}}}}{{{\rm{[1}}{\rm{.0]}}}}\; = \;2.218\,{\rm{V}}\]

Because Q = 2.5, the difference between E_cell and E°_cell is not very large due to the logarithm. So, at this concentration ratio, the cell potential is smaller than E°_cell, however, the reaction is still spontaneous and will proceed forward until an equilibrium is reached when Q = K and E_cell = 0. 

2) [Sn²⁺] = 3.0 M and [Mg²⁺] = 1.0 M

Part (a)

There is more reactant than product in the system, so the reaction will tend to shift forward more than it would under standard conditions. Therefore, E_cell should be larger than E°_cell: Q < 1, E_cell > E°_cell. 

Part (b)         

We have already determined the E°_cell, so we can use it in the Nernst equation to calculate the E_cell:       

\[E\; = {\mkern 1mu} {E^o}\; – {\mkern 1mu} \frac{{{\rm{0}}.{\rm{0257}}\;{\rm{V}}}}{{\rm{2}}}{\mkern 1mu} {\rm{ln}}{\mkern 1mu} \frac{{[{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}]}}{{[{\rm{S}}{{\rm{n}}^{{\rm{2 + }}}}]}}\]

\[E\; = \,2.23\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[1]}}}}{{{\rm{[3}}{\rm{.0]}}}}\; = \;2.244\,{\rm{V}}\]

Check Also

• Balancing Redox Reactions
• Galvanic Cells
• How to Calculate Standard Cell Potential
• The Correlation Between E^o_cell, ΔG°, and K
• Nernst Equation
• Concentration Cells
• Electrolytic Cells
• Electrolysis
• Electrolysis of Water
• Calculating the Mass of Metal in Electroplating
• Cell Potential Practice Problems
• E^o, ΔG^o, K – Practice Problems
• Electrochemistry Practice Problems