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 become oxidised.

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: $E_{\text{cell}}^{\ominus }=E_{\text{reduced}}^{\ominus }-E_{\text{oxidised}}^{\ominus }$

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

$E_{\text{cell}}^{\ominus }=E_{\text{reduced}}^{\ominus }-E_{\text{oxidised}}^{\ominus }$

Step 3: Substitution and correct evaluation

$E_{\text{cell}}^{\ominus }=+0.80-(+0.34)=+0.46\text{ V}$

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

$E_{\text{cell}}^{\ominus }=E_{\text{reduced}}^{\ominus }-E_{\text{oxidised}}^{\ominus }$

Step 3: Substitution and correct evaluation

$E_{\text{cell}}^{\ominus }=+1.36-(-2.71)=+4.07\text{ V}$

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

$E_{\text{cell}}^{\ominus }=E_{\text{reduced}}^{\ominus }-E_{\text{oxidised}}^{\ominus }$

Step 3: Substitution and correct evaluation

$E_{\text{cell}}^{\ominus }=+1.51-(+0.77)=+0.74\text{ V}$

Step 4: Predict feasibility

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

Limitations of using E⦵ for predictions

Predictions using standard electrode potentials have limitations and may not always accurately reflect the feasibility of reactions under non-standard conditions.

The key limitations are:

1. Non-standard conditions - Standard electrode potentials are determined at 1 mol dm^-3 concentrations, 298 K, and 100 kPa pressure. Deviations from these conditions, such as changes in concentration, temperature, or pressure, will affect the electrode potentials and may alter reaction feasibilty.
2. Unfavorable reaction kinetics - Standard potentials only consider thermodynamics, not kinetics. A reaction predicted to be feasible based on E^⦵ may still proceed very slowly if it has a high activation energy or slow reaction rate.
3. Non-aqueous reactions - Standard electrode potentials are determined for aqueous equilibria. They may not accurately predict feasibility for reactions in non-aqueous solvents.