2.6 Electrochemical Series | Electrochemistry | Chemistry 12

The Electrochemical Series (ECS) is a table of half-reactions arranged in order of increasing standard reduction potential ( $E^{\circ}$ ). It is one of the most powerful tools in electrochemistry for predicting the feasibility of redox reactions and the direction of electron flow.

Standard Electrode Potential ( $E^{\circ}$ )

The standard electrode potential ( $E^{\circ}$ ) is the potential (voltage) of a half-cell measured under standard conditions:

Temperature: 298 K (25 °C)
Ion concentration: 1 mol dm⁻³
Gas pressure: 100 kPa (1 atm)

All $E^{\circ}$ values are measured relative to the Standard Hydrogen Electrode (SHE), which is assigned a potential of exactly 0.00 V.

By convention, $E^{\circ}$ values are always quoted as reduction potentials: $M^{n +} (a q) + n e^{-} \to M (s) E^{\circ}$

Standard Cell Potential ( $E_{ce l l}^{\circ}$ )

The standard cell potential is the overall voltage produced by an electrochemical cell under standard conditions. It is calculated by combining the standard electrode potentials of the two half-cells:

$E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$

Where:

Cathode = the electrode where reduction occurs (higher $E^{\circ}$ )
Anode = the electrode where oxidation occurs (lower $E^{\circ}$ )

Example: For the Daniell cell (Zn–Cu cell):

$E^{\circ} (C u^{2 +} / C u) = + 0.34 V$ (cathode)
$E^{\circ} (Z n^{2 +} / Z n) = - 0.76 V$ (anode)

$E_{ce l l}^{\circ} = + 0.34 - (- 0.76) = + 1.10 V$

Arrangement of the Electrochemical Series

The ECS lists half-reactions from the most negative $E^{\circ}$ at the top to the most positive $E^{\circ}$ at the bottom:

Half-reaction	$E^{\circ}$ (V)
$L i^{+} (a q) + e^{-} \to L i (s)$	$- 3.05$
$K^{+} (a q) + e^{-} \to K (s)$	$- 2.92$
$M g^{2 +} (a q) + 2 e^{-} \to M g (s)$	$- 2.37$
$Z n^{2 +} (a q) + 2 e^{-} \to Z n (s)$	$- 0.76$
$F e^{2 +} (a q) + 2 e^{-} \to F e (s)$	$- 0.44$
$2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$	$0.00$
$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$	$+ 0.34$
$A g^{+} (a q) + e^{-} \to A g (s)$	$+ 0.80$
$F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q)$	$+ 2.87$

Key trends:

Species at the top (most negative $E^{\circ}$ ) are the strongest reducing agents — they readily lose electrons.
Species at the bottom (most positive $E^{\circ}$ ) are the strongest oxidising agents — they readily gain electrons.

Predicting Feasibility of a Reaction

A redox reaction is spontaneous (feasible) if: $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} > 0$

A positive $E_{ce l l}^{\circ}$ indicates the reaction proceeds spontaneously in the forward direction under standard conditions.

A negative $E_{ce l l}^{\circ}$ indicates the reaction is non-spontaneous in the forward direction.

Rule: In the ECS, a species with a higher (more positive) $E^{\circ}$ will oxidise a species with a lower (more negative) $E^{\circ}$ .

Direction of Electron Flow

Electrons always flow from the anode (lower $E^{\circ}$ , oxidation) to the cathode (higher $E^{\circ}$ , reduction) through the external circuit.

Example: In the Zn–Cu cell:

Zn is oxidised at the anode: $Z n (s) \to Z n^{2 +} (a q) + 2 e^{-}$
$C u^{2 +}$ is reduced at the cathode: $C u^{2 +} (a q) + 2 e^{-} \to C u (s)$
Electrons flow from Zn → Cu through the external wire.

Applications of the Electrochemical Series

1. Displacement of Hydrogen from Acids

Metals with $E^{\circ} < 0 V$ (above hydrogen in the ECS) can displace $H_{2}$ from dilute acids: $Z n (s) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} (g)$ Copper ( $E^{\circ} = + 0.34 V$ ) cannot displace hydrogen because its $E^{\circ}$ is more positive than that of $H^{+} / H_{2}$ .

2. Displacement of Metals from Salt Solutions

A metal higher in the ECS (more negative $E^{\circ}$ ) displaces a metal lower in the ECS (more positive $E^{\circ}$ ) from its salt solution: $C u (s) + 2 A g N O_{3} (a q) \to C u (N O_{3})_{2} (a q) + 2 A g (s)$ $E_{ce l l}^{\circ} = + 0.80 - (+ 0.34) = + 0.46 V > 0 ✓$

Worked Example

Question: Predict whether the reaction between $F e^{2 +}$ and $B r_{2}$ is spontaneous under standard conditions.

Given:

$E^{\circ} (B r_{2} / B r^{-}) = + 1.07 V$
$E^{\circ} (F e^{3 +} / F e^{2 +}) = + 0.77 V$

Solution:

$B r_{2}$ has the higher $E^{\circ}$ , so it acts as the oxidising agent (cathode).
$F e^{2 +}$ is oxidised to $F e^{3 +}$ (anode).

$E_{ce l l}^{\circ} = + 1.07 - (+ 0.77) = + 0.30 V$

Since $E_{ce l l}^{\circ} > 0$ , the reaction is spontaneous.

Overall equation: $2 F e^{2 +} (a q) + B r_{2} (a q) \to 2 F e^{3 +} (a q) + 2 B r^{-} (a q)$

Half-reaction

E^{\circ}

(V)

L i^{+} (a q) + e^{-} \to L i (s)

- 3.05

K^{+} (a q) + e^{-} \to K (s)

- 2.92

M g^{2 +} (a q) + 2 e^{-} \to M g (s)

- 2.37

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s)

- 0.76

F e^{2 +} (a q) + 2 e^{-} \to F e (s)

- 0.44

2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)

0.00

C u^{2 +} (a q) + 2 e^{-} \to C u (s)

+ 0.34

A g^{+} (a q) + e^{-} \to A g (s)

+ 0.80

F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q)

+ 2.87