Using E⦵ Values

This lesson covers: 

1. What the electrochemical series is
2. Using the electrochemical series to calculate standard cell potentials
3. Predicting whether reactions will occur using electrode potentials
4. The Nernst equation for non-standard conditions
5. The link between standard cell potential and ΔG

The electrochemical series ranks redox half-reactions

The electrochemical series is a list of reduction half-equations in order of their standard electrode potentials (E^⦵).

• The half-equation with the most positive E^⦵ value is placed at the bottom. The oxidised species in this half-equation is the strongest oxidising agent as it most readily accepts electron to become reduced.
• The half-equation with the most negative E^⦵ value is at the top. This reduced species in this half-equation is the strongest reducing agent as it most readily loses electrons to becom

For example, the half-equation for fluorine gas gaining electrons has a very positive E^⦵ value (+2.87 V):

F_2(g) + 2e^- ➔ 2F^-_(aq)             E^⦵ = +2.87 V

This tells us that fluorine has a great tendency to be reduced, so is a powerful oxidising agent.

By contrast, the half-equation for lithium metal losing an electron has a very negative E^⦵ value (-3.04 V):

Li⁺_(aq) + e^- ➔ Li_(s)                  E^⦵ = -3.04 V

This negative E^⦵ value indicates that the forward reaction is not favoured. Rather, lithium metal very readily undergoes oxidation to form lithium ions, so is a strong reducing agent.

Calculating cell potentials from E^⦵ values

Standard electrode potentials can be used to calculate the standard cell potential (E^⦵_cell) for an electrochemical cell:

1. Write the reduction half-reactions for both species.
2. Determine which reaction occurs in the oxidation direction (more negative E^⦵) and which occurs in the reduction direction (more positive E^⦵).
3. Combine the half-reactions to give the overall redox reaction.
4. Calculate E^⦵_cell using the equation:

Ecell⊖​=Ereduced⊖​−Eoxidised⊖​

Worked example 1 - Calculating E^⦵_cell for an electrochemical cell

Calculate E^⦵_cell for a cell comprising copper and silver ions, given the standard electrode potential data below. The overall reaction is:

Cu_(s) + 2Ag⁺_(aq) ➔ Cu²⁺_(aq) + 2Ag_(s)

Step 1: Identify oxidation and reduction half-reactions

The copper half-reaction proceeds in the oxidation direction (Cu_(s) ➔ Cu²⁺_(aq) + 2e^-), as indicated by its lower positive E^⦵ value relative to silver's half-reaction.

The silver half-reaction  proceeds in the reduction direction (Ag⁺_(aq) + e^- ➔ Ag_(s)), as indicated by its higher E^⦵ value.

Step 2: Equation

Ecell⊖​=Ereduced⊖​−Eoxidised⊖​

Step 3: Substitution and correct evaluation

Ecell⊖​=+0.80−(+0.34)=+0.46 V

Worked example 2 - Calculating E^⦵_cell for an electrochemical cell

Calculate E^⦵_cell for a cell comprising sodium metal and chlorine gas, given the standard electrode potential data below. The overall reaction is:

2Na_(s) + Cl_2(g) ➔ 2Na⁺_(aq) + 2Cl^-_(aq)

Step 1: Identify oxidation and reduction half-reactions

The sodium half-reaction proceeds in the oxidation direction (Na_(s) ➔ Na⁺_(aq) + e^-), as indicated by its lower (i.e. negative) E^⦵ value relative to chlorine's half-reaction.

The chlorine half-reaction proceeds in the reduction direction (Cl_2(g) + 2e^- ➔ 2Cl^-_(aq)), as indicated by its higher (i.e. positive) E^⦵ value.

Step 2: Equation

Ecell⊖​=Ereduced⊖​−Eoxidised⊖​

Step 3: Substitution and correct evaluation

Ecell⊖​=+1.36−(−2.71)=+4.07 V

Predicting reaction feasibility using electrode potentials

To predict if a redox reaction will occur spontaneously, compare the standard reduction potentials of the two half-reactions.

The half-reaction with the more negative E^⦵ will proceed in the oxidation direction, while the half-reaction with the more positive E^⦵ will proceed in the reduction direction.

If the resulting E^⦵_cell is positive, the reaction is feasible under standard conditions. The more positive the value of E^⦵_cell, the more feasible the redox reaction is.

Worked example 3 - Predicting the feasibility of a redox reaction

Calculate E^⦵_cell for the cell comprising of MnO₄^- ions and Fe²⁺ ions, given the following standard electrode potentials.

Determine whether the reaction is spontaneous under standard conditions.

Step 1: Compare standard reduction potentials

The half-reaction involving manganese has a more positive E^⦵ value, indicating it proceeds in the reduction direction:

MnO₄^-_(aq) + 8H⁺_(aq) + 5e^- ➔ Mn²⁺_(aq) + 4H₂O_(l)

The half-reaction involving iron has a less positive E^⦵ value, indicating it proceeds in the oxidation direction:

Fe²⁺_(aq) ➔ Fe³⁺_(aq) + e^-

Step 2: Equation

Ecell⊖​=Ereduced⊖​−Eoxidised⊖​

Step 3: Substitution and correct evaluation

Ecell⊖​=+1.51−(+0.77)=+0.74 V

Step 4: Predict feasibility

As E^⦵_cell is positive, the reaction between MnO₄^- and Fe²⁺ ions is spontaneous under standard conditions.

The Nernst equation for non-standard conditions

The Nernst equation relates the cell potential (E_cell) to the standard cell potential (E^⦵_cell) and concentrations of oxidised and reduced species at 298 K:

Ecell​=Ecell⊖​+(z0.059​)log10​[reduced species][oxidised species]​

Where:

• E_cell = cell potential under non-standard conditions (V)
• E^⦵_cell = standard cell potential (V)
• z = number of electrons transferred in the reaction
• [oxidised species] = concentration of the oxidised species (mol dm^-3)
• [reduced species] = concentration of the reduced species (mol dm^-3)

Worked example 4 - Calculating E_cell under non-standard conditions

Calculate the electrode potential at 298 K of a Zn²⁺/Zn half-cell, given the concentration of Zn²⁺ ions is 0.050 mol dm^-3.

The half-cell reaction is:

Zn²⁺_(aq) + 2e^- ⇌ Zn_(s)           E^⦵ = -0.76 V

Step 1: Identify oxidised and reduced species

The oxidised species is Zn²⁺ as it can gain electrons to be reduced to solid zinc.

The reduced species is solid Zn, as it represents the reduction product.

Step 2: Determine number of electrons transferred (z)

z = 2, as two electrons are transferred in this reaction.

Step 3: Nernst equation

Ecell​=Ecell⊖​+(z0.059​)log[reduced species][oxidised species]​

Given the concentration of solid Zn doesn’t change, the equation simplifies to:

Ecell​=Ecell⊖​+(z0.059​)log[oxidised species]

Step 4: Substitution and correct evaluation

Ecell​=(−0.76)+(20.059​)log[0.050]

Ecell​=(−0.76)+(−0.038)=−0.80 V

Relationship between standard cell potential and Gibbs free energy change

The standard cell potential (E^⦵_cell) is directly related to the standard Gibbs free energy change (ΔG^⦵) for a redox reaction by the equation:

ΔG⊖=−nFEcell⊖​

Where:

• ΔG^⦵ = standard Gibbs free energy change (J mol^-1)
• n = number of moles of electrons transferred in the balanced redox equation
• F = Faraday constant (96,500 C mol^-1)
• E^⦵_cell = standard cell potential (V)

The sign of ΔG^⦵ indicates the feasibility of the reaction under standard conditions:

• If ΔG^⦵ < 0 (E^⦵_cell > 0), the reaction is spontaneous
• If ΔG^⦵ > 0 (E^⦵_cell < 0), the reaction is not spontaneous

This relationship allows chemists to predict the feasibility of redox reactions using standard cell potentials, which can be calculated from standard reduction potentials.

Worked example 5 - Calculating ΔG^⦵

Calculate the standard Gibbs free energy change (ΔG^⦵) in kJ mol^-1 for the reaction:

2Al_(s) + 3Cu²⁺_(aq) ➔ 2Al³⁺_(aq) + 3Cu_(s)      E^⦵_cell = +2.00 V

The Farady constant, F = 96,500 C mol^-1.

Step 1: Determine number of moles of electrons transferred (n)

In this reaction, 6 moles of electrons are transferred as each aluminium atom donates 3 electrons and there are 2 aluminium atoms involved.

Step 2: Gibbs free energy equation

ΔG⊖=−nFEcell⊖​

Step 3: Substitution and correct evaluation

ΔG⊖=−6×96,500×2.00=−1,158,000 J mol−1

Step 4: Conversion of J mol^-1 into kJ mol^-1

To convert from J mol^-1 into kJ mol^-1, divide by 1,000

−1,158,000 J mol−1=−1,158 kJ mol−1

The negative value of ΔG^⦵, which is consistent with the positive standard cell potential, indicates that the reaction is spontaneous under standard conditions.