Cell Potential

Think of a battery as a small city divided into two regions; one side has extra electrons (negative), and the other side has spots that need electrons (positive). The voltage is the push that makes electrons flow from the negative to the positive side, like workers filling empty spots in the city. This difference in energy between the two areas is called cell potential, and it’s what powers devices like your remote or car engine. The greater the difference, the stronger the push for electrons to move, creating the electricity that powers things.

What is Cell Potential?

Cell potential, E_cell​, refers to the voltage difference between the two half-cells in an electrochemical cell. This variation is caused by the flow of electrons, which occurs during a redox reaction. In this reaction, one species undergoes oxidation (loses electrons), becoming positively charged, while another undergoes reduction (gains electrons), becoming negatively charged. The extent to which each species is predisposed to lose or gain electrons determines the overall cell potential.

Thus, the cell potential is a direct result of the difference in the tendency of the oxidising agent to accept electrons and the reducing agent to donate them. This potential is quantified in volts, providing a measurable value for the energy variation between the two half-cells that enables the electrochemical reaction to generate electrical current.

Cell Potential Formula,

The cell potential formula is widely used to find the electric potential of the cell. The diagram given below shows the cell potential formula :

Electrochemical Cell

An electrochemical cell is made up of two half-cells, with each having a metal electrode immersed in a solution of its respective metal ions. In one half-cell, oxidation occurs, while in the other half-cell, reduction takes place. These half-cells are connected by a wire that allows electrons to flow from one half-cell to the other.

• Let’s use zinc as the anode and copper as the cathode for this example. At the anode (the zinc side), the zinc metal undergoes oxidation by losing electrons.
• The zinc (Zn) atoms at the electrode lose two electrons and become zinc ions (Zn²⁺), then dissolve into the solution. These electrons then travel through the wire to the cathode (the copper side).
• At the cathode, the copper ions present in the solution (Cu²⁺) gain the electrons that traveled through the wire and get reduced. This means the copper ions (Cu²⁺) are transformed into solid copper (Cu), which deposits onto the copper electrode.
• To complete the circuit and maintain charge balance, the two half-cells are connected by a porous disk, which allows the flow of ions between the two solutions.
• The porous disk lets anions (negative ions) flow from the cathode side (where the solution becomes more positively charged as copper ions are reduced) to the anode side, where there’s an increasing concentration of zinc ions. This helps keep the charge balance in each half-cell as the reactions occur.
• Thus, the porous disk allows ions to flow and ensures that the electrochemical reactions can continue, generating the flow of electrons that powers devices like batteries.

How Electrochemical Potential Regulate Electron Transfer?

• The flow of electrons from the anode to the cathode in an electrochemical cell is regulate by the difference in electrochemical potential between the two electrodes.
• For electron transfer to occur, there must be a potential energy difference that makes the redox reactions thermodynamically favorable. This difference is a result of the tendency of the anode to undergo oxidation (electron loss) and the cathode to undergo reduction (electron gain).
• The anode possesses a greater electrochemical potential for oxidation, while the cathode has a lower potential for this process. Consequently, electrons are driven to flow from the anode, where the potential for oxidation is greater, to the cathode, where the potential for reduction is more favorable.
• This movement of electrons can be analogized to a mass moving from a higher energy state to a lower one, much like a rock descending from a cliff due to the difference in gravitational potential energy.
• The energy released by this flow of electrons is what powers the electrochemical cell. The magnitude of this potential difference between the anode and cathode is quantified as the cell potential E_cell, which determines the amount of energy available to drive the external circuit and perform work.

Note : The cell potential is the result of the difference in the reduction potentials between the cathode and the anode. It is calculated by subtracting the reduction potential of the anode from that of the cathode.

Mathematically, this is expressed as:

E_cell∘​=E ^∘​_red ,cathode​−E^∘_​red, anode

This difference in reduction potentials determines the overall voltage of the electrochemical cell, driving the electron flow from the anode to the cathode.

Types of Cell Potential

There are two main types of cell potential: 

1. Standard Cell Potential
2. Non-Standard Cell Potential

1. Standard Cell Potential

The standard cell potential E_cellrepresents the voltage generated by the electrochemical cell and is determined by the difference in the potentials of the two electrodes. To calculate this difference, we use the following equation:

E^∘_cell =E ^∘​_red ,cathode​−E^∘_​red, anode - (1a)

Where,

• E_cell​ is the standard cell potential, measured under conditions of 1M concentration, 1 bar pressure, and 298 K temperature.
• E ^∘​_red,cathode​ is the standard reduction potential for the reduction half-reaction occurring at the cathode.
• E^∘_​red, anode​ is the standard reduction potential for the oxidation half-reaction occurring at the anode.

The potential values are typically expressed in volts (V). Alternatively, this relationship can also be written as a sum of the reduction potential and oxidation potential, as shown below:

E^∘_cell =E ^∘​_red ,cathode​ +E^∘_​ox, anode - (1b)

Here, we shift the approach from subtracting the reduction potentials to adding the reduction potential of the cathode and the oxidation potential of the anode.

Since the oxidation potential is the negative of the reduction potential (E^∘_ox =−E^∘_red), both methods yield the same result.

2. Non-Standard Cell Potential

The non-standard cell potential refers to the cell potential when the conditions deviate from the standard state (1 M concentration for solutions, 1 bar pressure for gases, and 298 K temperature). In non-standard conditions, factors such as the concentration of ions in solution, partial pressure of gases, and temperature affect the overall cell potential.

The Nernst equation is used to calculate the non-standard cell potential:

E_cell​ = E^∘_cell −0.0592/ n. ​log ( [ Red] _cathode /[ Ox] _cathode × [ Ox] _{anode /} [ Red] _anode)

Where,

• E_cell​ is the non-standard cell potential.
• E^∘_cell​ is the standard cell potential.
• n is the number of moles of electrons transferred in the reaction.
• [Red] and [Ox] are the concentrations of the reduced and oxidized forms of the species involved at the cathode and anode, respectively.

This equation accounts for how changes in concentration (or pressure, for gases) affect the cell potential, showing that the voltage of the cell will vary depending on the actual concentrations of the ions involved in the redox reaction.

For example, if the concentration of reactants is higher or the products' concentration is lower than standard conditions, the cell potential will be higher than the standard potential. Conversely, the cell potential will decrease if the reverse is true.

Application of cell potential

• Batteries: Cell potential facillitates energy storage and release in Lead-Acid, Lithium-Ion, and Nickel-Cadmium batteries.
• Fuel Cells: Cell potential is used to generate electricity through Hydrogen and Methanol Fuel Cells in vehicles and portable devices.
• Corrosion Prevention: Cell potential helps in Galvanic Corrosion and Cathodic Protection to prevent metal deterioration.
• Electroplating: Cell potential enables metal deposition in gold, silver, and chrome electroplating for industrial and decorative purposes.
• Electrolysis: Cell potential drives processes like water splitting, metal refining, and metal extraction.
• Electrochemical Sensors: Cell potential is used in blood glucose meters, pH meters, and environmental monitoring sensors.
• Energy Storage: Cell potential is central to technologies such as supercapacitors and redox flow batteries for efficient energy storage and release.
• Electrochemical Synthesis: Cell potential is applied in organic synthesis and water treatment for chemical production and purification.

Solved Examples

Example 1: Calculate the standard cell potential for the reaction:

Zn(s)+Cu²⁺(aq)→Zn²⁺(aq)+Cu(s)

Given,

Zn²⁺(aq)+2e−→Zn(s),E^∘=−0.76V

Cu²⁺(aq)+2e−→Cu(s),E^∘=+0.34V

Solution:

Identify the Half-Reactions and Their Standard Reduction Potentials
Zinc half-reaction (Anode)

Zn²⁺(aq)+2e−→Zn(s),E^∘=−0.76V

Copper half-reaction (Cathode):

Cu²⁺(aq)+2e−→Cu(s),E^∘=+0.34V

Write the Cell Reaction

Zn(s)+Cu²⁺(aq)→Zn²⁺(aq)+Cu(s)

⁛ Zinc (Zn) is oxidized at the anode, and copper ions (Cu²⁺) are reduced at the cathode.

Calculate the Standard Cell Potential

E^∘_cell=E^∘_cathode −E^∘_anode

E°_cathode (for Cu²⁺/Cu) = +0.34 V

E°_anode (for Zn²⁺/Zn) = -0.76 V

E^∘_cell=0.34 V−(−0.76 V)

E^∘_cell=1.10V

Example 2:Calculate the standard cell potential for the reaction:

Ni(s)+2Ag⁺(aq)→Ni²⁺(aq)+2Ag(s)

Solution:

Identify the Half-Reactions and their Standard Reduction Potentials

Silver half-reaction (Cathode)

Ag⁺⁽aq)+e−→Ag(s),E^∘=+0.80V

Nickel half-reaction (Anode)

Ni²⁺(aq)+2e−→Ni(s), E^∘=−0.23 V

Step 2: Write the Cell Reaction

Ni(s)+2Ag+(aq)→Ni²⁺⁽aq)+2Ag(s)

Nickel (Ni) is oxidized at the anode, and silver ions (Ag⁺) are reduced at the cathode.

Step 3: Calculate the Standard Cell Potential

E^∘_cell=E^∘_cathode −E^∘_anode

E°cathode (for Ag⁺/Ag) = +0.80 V

E°anode (for Ni²⁺/Ni) = -0.23 V

E^∘_cell=0.80 V−(−0.23 V)=1.03V

Example 3: An example of cell potential can be demonstrated using the redox reaction between zinc (Zn) and copper (Cu) in an electrochemical cell. The overall reaction can be written as:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

The standard reduction potential of Cu²⁺/Cu is +0.34 V, while the standard reduction potential of Zn²⁺/Zn is -0.76 V. Therefore, the Cu²⁺/Cu half-reaction will act as the cathode half-reaction and the Zn²⁺/Zn half-reaction will act as the anode half-reaction.

Solution:

The cell potential can be calculated using the formula:

E_cell = E°_cathode - E°_anode

E°_cathode = +0.34 V (Cu²⁺/Cu)
E°_anode = -0.76 V (Zn²⁺/Zn)

Ecell = +0.34 V - (-0.76 V) = +1.10 V

The positive value of the cell potential indicates that the reaction is spontaneous in the forward direction, i.e., Zn(s) is oxidized and Cu²⁺(aq) is reduced. The reaction proceeds until the concentrations of the reactants and products reach equilibrium.

To measure the actual cell potential under non-standard conditions, the Nernst equation can be used. For example, if the concentrations of Cu²⁺ and Zn²⁺ are both 1.00 M and the cell is at 25°C, the cell potential can be calculated as,

E_cell = E°_cell - (RT/nF) ln(Q)

Q = [Cu⁺²]/[Zn⁺²] = 1.00/1.00 = 1.00

Plugging in the values, we get

E_cell = +1.10 V - (0.0257 V) ln(1.00) = +1.10 V

Thus, the cell potential is the same as the standard cell potential under these conditions.

Example 4: Consider the following reaction:

Cu(s) + Ag⁺(aq) -> Cu²⁺(aq) + Ag(s)

Standard reduction potentials are,

Cu²⁺(aq) + 2e^- -> Cu(s)                (E° = +0.34 V)

Ag⁺(aq) + e^- -> Ag(s)                  (E° = +0.80 V)

A) Write the balanced equation for the cell reaction and calculate the standard cell potential. 

B) If the initial concentrations of Cu²⁺ and Ag⁺ are both 0.1 M and the cell operates at standard conditions, calculate the cell potential.

C) If the initial concentration of Cu²⁺ is 1 M and the initial concentration of Ag⁺ is 0.1 M, calculate the cell potential.

Solution:

Given:

The reaction is:

Cu(s)+Ag⁺(aq)→Cu²⁺(aq)+Ag(s)

Standard reduction potentials:

Cu²⁺(aq)+2e−→Cu(s) ,E^∘=+0.34V

Ag⁺(aq)+e−→Ag(s) ,E^∘=+0.80V

Part A: Balanced Equation and Standard Cell Potential

Balanced Equation:
The half-reactions for the reduction and oxidation:

For copper (Cu): Cu is oxidized from Cu(s) to Cu²⁺, so it is the anode (oxidation).

For silver (Ag): Ag⁺ is reduced to Ag(s), so it is the cathode (reduction).

The overall balanced reaction is:

Cu(s)+2Ag⁺(aq)→Cu²⁺(aq)+2Ag(s)

The standard cell potential is calculated using the formula:

E^∘_cell=E^∘_cathode∘−E^∘_anode

Thus,

E^∘_cell=+0.80 V−(+0.34 V)=+0.46 V

Part B: Standard Conditions (Cu²⁺ and Ag⁺ concentrations both 0.1 M)

E_cell=E^∘_cell=+0.46 V

Part C: Non-Standard Conditions (Cu²⁺ = 1 M and Ag⁺ = 0.1 M)

Reaction Quotient (Q):
Q=[Ag⁺][Cu²⁺]​=0.11​=10

Nernst Equation:
The Nernst equation is given by:Ecell=Ecell∘−0.059nlog⁡Q

Where,

E^∘_cell=+0.46 V(calculated above)

n=2n (since 2 electrons are transferred)

Q=10

Now, substituting the values

E_cell=+0.46 V−0.059/2log⁡10

Since log⁡10=1

E_cell=+0.46 V−0.059/2×1=+0.46 V−0.0295 V=+0.4305V

Example 5: Consider the following redox reaction, 2 Fe³⁺(aq) + 2 I^-(aq) → 2 Fe²⁺(aq) + I₂(s). Calculate the cell potential.

Solution:

To calculate the cell potential, we need to find the reduction potentials for each half-reaction and then subtract the reduction potential of the anode from the reduction potential of the cathode.

Half-reactions for the given reaction are,

Fe³⁺(aq) + e^- → Fe²⁺(aq)               E° = +0.77 V

I₂(s) + 2 e^- → 2 I^-(aq)                     E° = +0.54 V

Reduction Potential of the anode is the reduction potential of the species being oxidized, which is Fe³⁺ in this case. So, the reduction potential of the anode is +0.77 V.

The reduction potential of the cathode is the reduction potential of the species being reduced, which is I₂ in this case. So, the reduction potential of the cathode is +0.54 V.

The cell potential (E_cell) is the difference between the reduction potentials of the cathode and anode:

E_cell = E_cathode - E_anode
E_cell = +0.54 V - (+0.77 V)
E_cell = -0.23 V

Therefore, the cell potential for this reaction is -0.23 V,this means that the reaction is not spontaneous and the electrons will not flow from the anode to the cathode without an external energy source.

Related Reads,

• Electrochemical Cells
• Electrochemical Series
• Galvanic Cells
• Redox Reactions and Electrode Processes