**What is the Nernst Equation Used For?**

Nernst equation 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:

Both equations are applicable to reactions carried out at **25 ^{o}C (298 K)**, so let’s why is that the case, and how the equation is derived before looking at some examples.

**Deriving the Nernst Equation**

There are two equations we need to remember to derive the Nernst equation. One is the equation correlating the Gibbs free energy change under standard (Δ*G*°) and nonstandard conditions(Δ*G*) , and the reaction quotient (*Q*):

*where ΔG is the free-energy change under nonstandard-state conditions, ΔG° is the free-energy change under standard-state conditions, R is the gas constant, T is the temperature in kelvin, and Q is the reaction quotient*

*We talked about this equation in detail here, so feel free to check it out before going forward. *

* *

The second equation is the correlation between the change in Gibbs free energy and the cell potential:

Δ*G*° = –*nFE*°_{cell, }Δ*G* = –*nFE*_{cell}

_{ }

Combing these two equations, we can write that:

–*nFE*_{cell = }–*nFE*°_{cell }+ *RT*ln *Q*

* *

To simplify the equation, we divide each side by –*nF *which leads to the Nernst equation:

\[E\; = \,{E^o}\; – \,\frac{{RT}}{{nF}}\,\ln \,Q\]

Switching from the natural log, we can also write it as:

_{ }

\[E\; = \,{E^o}\; – \,\frac{{2.303RT}}{{nF}}\,\log \,Q\]

_{ }_{ }

*E is the cell potential in the given nonstandard condition, E ^{o} is the cell potential at standard conditions, n is the moles (coefficient) of electrons in the half-reactions, F is the Faraday constant (96,485 coulombs/mol e*

^{_}*)*

*and it is equivalent to the charge on one mole of electrons, R is the universal gas constant (*

*8.314*

*J / mol·K*

*), and Q is the reaction quotient*

_{ }

Notice that all the parameters in the formula are constant except for the temperature, and if we apply the equation to a reaction carried out at 298 K, we can get the most common ways of the Nernst equation:

\[E\; = \,{E^o}\; – \,2.303\, \times \,\frac{{8.314\,\frac{{\cancel{{\rm{J}}}}}{{\cancel{{{\rm{mol}}\,{\rm{K}}}}}}\; \times \;298.15\;\cancel{{\rm{K}}}}}{{\frac{{{\rm{n}}\;\cancel{{{\rm{mol}}\;{{\rm{e}}^{\rm{ – }}}}}}}{{\cancel{{{\rm{mol}}}}\,}}\;\left( {96,485\frac{{\cancel{{\rm{J}}}}}{{{\rm{V}}\,\cancel{{{\rm{mol}}\,{{\rm{e}}^{\rm{ – }}}}}\,}}} \right)}}\,\log \,Q\]

** **

**The Correlation Between ***E*_{cell} and *Q*

*E*

_{cell}and

*Q*

Looking at the Nernst equation for reactions at 25 ^{o}C, we can see 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
, it means [reactant] > [product], ln*Q*< 1*Q*< 0, so.*E*_{cell}>*E*°_{cell} - When
, it means [reactant] = [product], ln*Q*= 1*Q*= 0, so.*E*_{cell}=*E*°_{cell} - When
, it means [reactant] < [product], ln*Q*> 1*Q*> 0, so.*E*_{cell}<*E*°_{cell} - When
, it means [product] >> [reactant], ln*Q*=*K**Q*=*E*°_{cell}, so.*E*_{cell}= 0

So, we can divide the operation of the cell into four stages:

When the **Q < 1,** it means there is more of the reactants than products, and according to Le Châtelier’s principle, the reaction is going to have a greater tendency to **shift forward **to **form more of the products**. This, in turn, means the overall cell potential, at the given concentrations, is greater than the cell potential under standard conditions (*E*_{cell}** > E°_{cell}**).

At some point, as the concentration of the products is increasing and that of the reactants is decreasing, *Q* equals, the log term is zero, and the cell potential is the same as it would be under standard conditions: *E*_{cell} = *E***°**_{cell}* as Q *

**= 1**, and log 1 = 0.

As the reaction continues, Q becomes larger than 1, ** Q > 1**, because there is more product than reactant in the system, resulting in a smaller cell potential compared to that of under standard conditions (

*E*

_{cell}**<**). The reaction is still occurring, however, because of the smaller potential, it can do less work. Finally, when

*E*°_{cell}**, an**

*Q*equals*K***equilibrium**is reached and

**which means the**

*E*_{cell }becomes zero**cell is dead**and no more work can be done.

**What Goes in the Expression of ***Q*?

*Q*?

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*) + Pb^{2+}(*aq*) → Mn^{2+}(*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 – Examples**

A voltaic cell runs the following redox reaction:

Sn^{2+}(*aq*) + Mg(*s*) → Sn(*s*) + Mg^{2+}(*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 [Sn^{2+}] = 1.0 *M*and [Mg^{2+}] = 2.5 *M*

2) when [Sn^{2+}] = 3.0 *M*and [Mg^{2+}] = 1.0 *M*

**1) **[Sn^{2+}] = 1.0 *M*and [Mg^{2+}] = 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, Mg^{2+ }is higher than that of Sn^{2+ }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:

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

**F****irst, 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.

** **

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

Mg(*s*) → Mg^{2+}(*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)** [Sn^{2}^{+}] = 3.0 *M* and [Mg^{2+}] = 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}:

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

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

#### Practice

A voltaic cell runs the following redox reaction at 25 ^{o}C:

Fe^{2}^{+}(*aq*) + Mg(*s*) → Mg^{2+}(*aq*) + Fe(*s*)

Determine whether *E*_{cell} is larger or smaller than *E*°_{cell} and determine the *E*_{cell} for the following concentrations:

a) when [Fe^{2}^{+}] = 1.0 *M* and [Mg^{2+}] = 2.5 *M*

b) when [Fe^{2}^{+}] = 3.0 *M* and [Mg^{2+}] = 1.0 *M*

Calculate *E*°, *E*, and Δ*G *for the following cell reaction:

Mn(*s*) + Pb^{2+}(*aq*) → Mn^{2+}(*aq*) + Pb(*s*)

[Mn^{2+}] = 0.164 *M*, [Pb^{2+}] = 0.038 *M *

A galvanic cell operates on the following redox reaction:

2MnO_{4}^{–}(*aq*) + 10Br^{–}(*aq*) + 16H^{+}(*aq*) → 2Mn^{2+}(*aq*) + 5Br_{2}(*l*) + 8H_{2}O(*l*)

Calculate the cell potential at 25 ^{o}C, given the concentrations of the aqueous components are: [MnO_{4}^{–}] = 0.0250 *M*, [Br^{–}] = 0.0150 *M*, [Mn^{2}^{+}] = 0.380 *M*, and [H^{+}] = 0.64 *M*.

A voltaic cell consists of a Mg/Mg^{2}^{+} half-cell and a Sn/Sn^{2}^{+ }half-cell at 25 °C. The initial concentrations of Sn^{2}^{+ }and Mg^{2}^{+ }are 1.80 *M* and 0.250 *M*, respectively.

a) Calculate the initial cell potential.

b) What is the cell potential when the concentration of Sn^{2}^{+} dropped to 0.650 *M*?

c) Determine the concentrations of Sn^{2}^{+ }and Mg^{2}^{+ }ions when the cell potential falls to 2.15 V.

A voltaic cell consists of a Zn/Zn^{2}^{+} half-cell and a H_{2}/H^{+ }half-cell at 25 °C. Calculate the initial cell potential if [Zn^{2}^{+}] = 0.0250 *M*, [H^{+}] = 2.80 *M*, P(H_{2}) = 0.45 atm.

Determine the emf of a cell consisting of a Sn^{2+}/Sn half-cell and a Pt/H^{+}/H_{2} half-cell if [Sn^{2+}] = 0.12 *M*, [H^{+}] = 0.0650 *M*, and *P*(H_{2}) = 1.25 atm?