Cell Potential and Free Energy

While you are reading this, billions of neurons in your body are all working together to help make you a person. These nerves use chemistry and metals like sodium and potassium to operate. Now, what does a nerve cell, an electric car, and the root of a dandelion have in common? Well, if you thought of electrochemistry, then you would be right! All three of these things (and lots more) rely on cell potential and free energy to occur. How do they relate to each other? Stick around and use the electrochemistry in your body to find out!

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Jetzt kostenlos anmeldenWhile you are reading this, billions of neurons in your body are all working together to help make you a person. These nerves use chemistry and metals like sodium and potassium to operate. Now, what does a nerve cell, an electric car, and the root of a dandelion have in common? Well, if you thought of electrochemistry, then you would be right! All three of these things (and lots more) rely on cell potential and free energy to occur. How do they relate to each other? Stick around and use the electrochemistry in your body to find out!

- First, we will discuss the relationship between these parameters and what this means.
- Then, we will discuss how to calculate these terms.
- After, we will discuss changes in cell potential and Gibbs free energy.
- Finally, we will go over an example to help you practice.

In animals, nerve cells work by creating a **potential difference** of positive and negative ions. On one side of the cell there are accumulated Na^{+} ions, and on the other are a lot of K^{+} ions. The separation of these two ions creates a potential difference in the cell. So, when the cell opens channels, and allows this to happen, it produces energy. This is enough energy to stimulate the nerve cell which is what leads to anything you do; whether it is reading this article, sitting down, or even just breathing.

**Potential difference** is the amount of potential energy stored between two cells. The energy released when these cells can combine represents the maximum potential energy between the two.

A similar method is used in plant roots, which take up water and nutrients from the soil. A potential difference can be created within the root that allows stuff from the soil to flow into the roots. This concept is also what powers electric cars. In a battery, a potential difference generates power, which is then used to operate the device. This concept of potential difference, when applied to chemical cells, is called **cell potential**. When electrons flow from a place of high electron density, to a place of low electron density, they generate an electric charge, measured in Volts. This is referred to as a spontaneous reaction and has a negative free energy change.

To better understand the relationship between Gibbs free energy and cell potential, it is necessary that we recap spontaneous and nonspontaneous reactions. Recall that a spontaneous reaction will occur on its own with no extra help, while a nonspontaneous reaction will not occur unless something assists it.

What is the difference between cell potential and electrochemical cell potential? And how do they both relate to Gibbs free energy? Well, the term cell potential is more of an umbrella term that can be used to describe the potential difference in any kind of cell; whether that be biological or artificial. Electrochemical cell potential only describes the potential difference in an artificial cell. Essentially, one that's found in a lab or a battery.

Typically, we just use cell potential to describe everything, but it is important to recognize the slight difference between the two. They both relate to Gibbs free energy in the same way, since they can technically both describe the same thing. Quantifying this relationship is typically done with electrochemical cells, and to accomplish that, we need to know how to calculate cell potential.

In the study of Electrolysis, the potential difference between two half-cells is measured with a voltmeter. A positive voltage means that electrons are freely flowing from one cell to another. This means that a positive cell potential corresponds to a spontaneous reaction. For Example:

As electrons move from the Zn anode to the Cu cathode, a number appears on the voltmeter, which is the cell potential, *E**°*. We know that this reaction is spontaneous because it is proceeding without any external help. But we don't need to set up a galvanic cell every time we want to see if a cell potential is positive. It is easier to just calculate the cell potential based on the Standard Potential of each cell using our trusty formula.^{1}

$$ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} $$

So, to determine the standard cell potential of the cell, we subtract the standard potential of the anode by the standard potential of the cathode. In this reaction, zinc is giving electrons (oxidation) to copper, which is receiving electrons (reduction).

\begin{align} Cu^{2+} + 2e^- \rightarrow Cu(s) \qquad &E^{\circ} = 0.34~V \\ Zn^{2+} + 2e^- \rightarrow Zn(s) \qquad &E^{\circ} = -0.76~V \end{align}

Keep in mind that half-cell potentials are always written as reduction potentials (reduced form on the right of the arrow).

By combining these two half-cell potentials, we can determine the overall cell potential.

\begin{align} &E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \\ &E^{\circ}_{cell} = E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Zn^{2+}/Zn} \\ &E^{\circ}_{cell} = 0.34~V - (-0.76~V) \\ &E^{\circ}_{cell} = 1.10~V\end{align}

The overall cell potential of this voltaic cell is +1.10 V, and is thus a spontaneous reaction. But, how do we know which will be the cathode, and which will be the anode? Well, technically, we can pick whatever we want! Try calculating this reaction in the reverse direction—is it still a spontaneous reaction?

For a reaction to occur, there must be a positive cell potential. However, that doesn't mean we can't make a nonspontaneous reaction occur. Electrolysis shows us that if we want to make a nonspontaneous reaction happen, we just apply voltage until the cell potential becomes positive. It's as simple as that! That means that any reaction can be made possible with enough voltage.

Did you know: The standard electrode potential, E°—for each half reaction—is an *intensive* quantity. This means that it is not affected by the amount of substance in the cell. No matter how much you put in the cell, the value for your standard electrode potential will be the same.

We know now that cell potential and Gibbs' free energy change both evaluate the spontaneity of a reaction.

Spontaneous | Nonspontaneous | |

Cell Potential | E° > 0 | E° < 0 |

Gibbs Free Energy | Δ G°< 0 | Δ G° > 0 |

So when the cell potential is greater than 0, the reaction will be spontaneous. When Gibbs Free Energy is less than 0, the reaction is spontaneous. That means that cell potential and free energy are inversely proportional.

It is now pretty simple to see when a reaction will be spontaneous or not. We can also use one parameter to help us determine another—with math!

If we further break down the cell potential, it really looks like this:

$$ E = \frac {potential~difference~(J)} {charge~(C)} $$

Hint: J = Joule, a unit of energy. C = Coulomb, a measure of charge.

We can further simplify this by classifying that potential difference = *w* (work) and charge = *q*. This gives us:

$$ E = \frac {w} {q} $$

**Work** (w) represents an amount of energy transferred by the system to its surroundings, or by the surroundings into the system. Imagine a cup of hot coffee. As time passes, the coffee cup gets colder. If the coffee cup is the system, and everything else are the surroundings. The coffee cup is putting thermal energy into its surroundings. The system is doing work on the surroundings. This continues until the coffee cup and the surroundings are the same temperature.

If we express this mathematically, the system will be **decreasing** in energy. So, it will have a **negative** value since energy is leaving the system. In this example of potential difference, it is the maximum amount of work that can be done to the surroundings and will correspond to a negative energy value.

By rearranging the formula and considering the work function, we arrive at:

$$ w = -qE^{\circ}_{cell} $$

If we remember that Faraday's Law is related to charge and number of electrons (*n*), we can substitute it into our *q *function. Recall **Faraday's Constant**, which represents the charge of electrons (in C) per mole**:**

$$ F=96485~C~mol^{-1} $$

Hint: A common way to write units is in this way. The negative exponent means that it is in the denominator. It can also be written like this:

$$ C~mol^{-1}=C\cdot mol^{-1} = \frac{C} {mol} $$

It is important that you can recognize the different ways in which units might be listed.

\begin{align} q &= nF \\ w &= -qE^{\circ}_{cell} \\ w &= -nFE^{\circ}_{cell} \end{align}

Now that we've come to our work function, we can relate this to our free energy. When dealing with the maximum amount of work on a system, we can exchange the term for Gibbs Free Energy. Since Δ*G* = *w*_{max}, we get:

$$ \Delta G^{\circ} = -nFE^{\circ}_{cell} $$

With this new formula, it is now fairly simple to determine the cell potential when we know the Gibbs free energy change and vice versa.

Now that we have a formula to change between cell potential and free energy, we can easily determine one from the other. If we know the values of one, it is a lot easier to calculate the others than it is to measure them.

Now that we know if a reaction will be spontaneous, we also would like to know whether it is at Equilibrium or not. We know that the process of Calculating Equilibrium Constant relates it to our Gibbs free energy.

$$ \Delta G = -RT \ln{K} $$

With our previous derivation, we can now relate the equilibrium constant to our cell potential.

\begin{align} -nFE^{\circ}_{cell} &= -RT\ln{K} \\ E^{\circ}_{cell} &= \frac {RT} {nF} \ln{K} \end{align}

With this formula, we can not only measure the spontaneity of a reaction, but we can also measure its progress.

If we assume that we are under standard conditions we can quickly simplify this.

Let's assign the values for our standard conditions:

*T*(temperature) = 25 °C = 273.15 K (kelvin)*R*(ideal gas constant) = 8.314 J mol^{-1}K^{-1}*F*= 96485 C mol^{-1}

By plugging these values in, we end up with:

$$ E^{\circ}_{cell} = \frac {0.0257~V} {n} \ln{K} $$

Determining changes in cell potential just became easier!

Since we have a new parameter, equilibrium, we should understand what this is telling us. Well, we know that reactions are typically under Dynamic Equilibrium, where they will try to reach a steady state. Atoms and molecules want to be at equilibrium because being higher in energy is hard work!

If we think about spontaneous reactions, we now know that it means that the forward reaction is probable, while the reverse reaction is improbable. How does that tie into equilibrium? Well, in a Dynamic Equilibrium, the equilibrium is always shifting back and forth slightly. In a spontaneous reaction, this means that it is more favored to go in the forward direction. So while it is always shifting, it is shifting more forward than it is backward.

Perhaps this table will help you visualize this.

Spontaneous | Nonspontaneous | |

Cell Potential | E°_{cell} > 0 | E°_{cell} < 0 |

Gibbs Free Energy | ΔG° < 0 | ΔG° > 0 |

Equilibrium Constant | K > 1 | K < 1 |

Now that we have derived the equation relating Gibbs free energy and cell potential, it should be noted that we have been dealing with the "standard" cell potential. Standard conditions are always denoted with a degree symbol (°). So, when you have seen *E°*_{cell}, or just *E°*, that means that the cell is under standard conditions.

If there is no degree symbol indicated, such as *E*, then you can assume the cell is not under standard conditions. This could be anything, such as a change in cell conditions, or if we are dealing with an electrolytic cell, which requires a voltage to operate.

We have seen what cell potential and free energy are and how to calculate them. Let's take a look at an example to bring this all to life.

A piece of sodium metal is placed in an electrochemical cell containing water. Calculate Δ*G* from the standard cell potentials.

$$ Na(s) + 2H_2O(l) \rightarrow H_2(g) + 2OH^-(aq) + Na^+(aq) $$ \begin{align} 2H_2O(l) +2e^- &\rightarrow H_2(g) + 2OH^- &E^{\circ}_{cell} = -0.41~V \\ [Na^+(aq) + e^- &\rightarrow Na(s)]\times 2 &E^{\circ}_{cell} = -2.71~V \end{align}

To be able to combine our two half reactions, we must make the electrons cancel out. This can be done by multiplying our sodium half reaction by 2. But we don't multiply the standard half cell potential by 2. Can you remember why?

\begin{align} 2Na(s) + 2H_2O(l) +2e^- &\rightarrow H_2(g) +2OH^-(aq) + 2Na^+(aq) + 2e^- \\ 2Na(s) + 2H_2O(l) &\rightarrow H_2(g) +2OH^-(aq) + 2Na^+(aq) \end{align}

In the overall reaction, sodium metal is being oxidized, so it is serving as the anode while water serves as the cathode. Now that we know each, we can calculate the overall cell potential.

\begin{align} &E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \\ &E^{\circ}_{cell} = -0.41~V - (-2.71~V) \\ &E^{\circ}_{cell} = 2.30~V \end{align}

Now we can use our cell potential to determine our Gibbs free energy.

$$ \Delta G^{\circ} = -nFE^{\circ}_{cell} $$

For this reaction, *n* = 2, since there were 2 electrons in our half reactions.

$$ \Delta G^{\circ} = -2(96485~C~mol^{-1})(2.30~V) $$

For our units to cancel out, we will need to express our cell potential in \(J~C^{-1}\)

Remember that: \(cell~potential = \frac {potential~difference~(J)} {charge~(C)}\)

\begin{align} &\Delta G^{\circ} = -2(96485~C~mol^{-1})(2.30~J~C^{-1}) \\ &\Delta G^{\circ} = -4.44 \times 10^5~J~mol^{-1}(\frac {1~kJ} {1000~J}) \\ &\Delta G^{\circ} = -444~kJ~mol^{-1} \end{align}

Is this reaction spontaneous, or not? If you answered yes, then you would be correct! Good job!

**Cell Potential**is the potential energy difference between two half cells, measured in Volts- A
**positive**cell potential value means the reaction will be spontaneous, while a**negative**Gibbs free energy means a reaction is spontaneous - We can relate cell potential with Gibbs free energy with a formula:\( \Delta G^{\circ} = -nFE^{\circ}_{cell}\)
- Relating cell potential with
**equilibrium**is also possible:\( E^{\circ}_{cell} = \frac {RT} { nF} \ln {K}\) - By assuming standard temperature & pressure (
**STP**), we can shorten the formula to:\( E^{\circ}_{cell} = \frac {0.0257~V} { n} \ln {K}\)

- Nivaldo Tro, Travis Fridgen, Lawton Shaw,
*Chemistry a Molecular Approach*,*3rd ed*., 2017 - Daniel Nocera, Artificial Photosynthesis at Efficiencies Greatly Exceeding That of Natural Photosynthesis,
*Accounts of Chemical Research*, 2019

The equilibrium constant can be calculated from cell potential with the following formula:

*E°* = *(RT/nF)* ln *K*

Gibbs' free energy and cell potential both measure whether a reaction is thermodynamically allowed (spontaneous). They can be mathematically related with the following formula:

Δ*G* = *-RT* ln *K*

A cell potential of *E° = 1.10 *V represents what?

A spontaneous reaction.

Which reaction will be spontaneous?

*E° *< 0

Δ*G* > 0

*K* > 1

You are presented with the following reaction. What will be the overall cell potential?

Na(s) + 2 H2O(l) ---> H2(g) + 2 OH-(aq) + Na+(aq)

Here are the two half reactions:

[ Na+ + e- ---> Na(s) ] *E°* = -0.41 v

[ 2 H20(l) + 2e- ---> H2(g) + 2 OH- ] *E°* = -2.71 V

*E° *= -2.3 V

In a Galvanic (Voltaic) cell electrons move in which direction? Select multiple.

From the cathode to the anode

What is the formula for converting between cell potential and Gibbs free energy under standard conditions?

Δ*G *= -*RT* ln *K*

What is the value of Faraday's constant?

96485 C mol^{-1}

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