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.