Electrolysis

Electrode Potential ($E$)

When a metal rod is immersed in an aqueous solution containing its own ions (e.g., a zinc rod placed in a $\text{ZnSO}_4$ solution), an electrical potential difference is established between the metal strip and the surrounding liquid. This potential difference is known as the Electrode Potential.

Two opposing processes take place simultaneously when a metal electrode is placed in an electrolyte solution containing its ions:

Oxidation Tendency (Solution Pressure): Metal atoms on the electrode lose electrons and dissolve into solution as positive cations ($\text{M} \rightarrow \text{M}^{n+} + n e^-$). This leaves a net negative charge on the electrode.
Reduction Tendency (Osmotic Pressure): Metal cations in solution gain electrons from the electrode and deposit as neutral metal atoms ($\text{M}^{n+} + n e^- \rightarrow \text{M}$). This leaves a net positive charge on the electrode.

At equilibrium, an electrical double layer (Helmholtz double layer) forms at the metal-electrolyte interface, giving rise to the electrode potential.

Standard Electrode Potential ($E^\circ$)

Definition: The Standard Electrode Potential ($E^\circ$) of a half-cell is the potential difference established between a metal electrode and a solution containing its ions at a standard concentration of $1.0\text{ M}$ ($\text{1 mol/dm}^3$), measured at $25^\circ\text{C}$ ($298\text{ K}$) and $1\text{ atm}$ pressure, referenced against the Standard Hydrogen Electrode (SHE).

Standard Hydrogen Electrode (SHE):

By IUPAC convention, the Standard Hydrogen Electrode is assigned an arbitrary potential value of exactly $0.00\text{ Volts}$ ($E^\circ = 0.00\text{ V}$). All standard reduction potentials are measured relative to this reference half-cell.

Standard Reference Reaction:    2H⁺_{(aq, 1M)} + 2e⁻ ⇌ H₂_{(g, 1 atm)}     E° = 0.00 V

Electromotive Force of a Cell ($E_{\text{cell}}$)

Definition: The Electromotive Force (EMF) of an electrochemical cell (denoted as $E_{\text{cell}}$ or $E^\circ_{\text{cell}}$ under standard conditions) is defined as the maximum potential difference between the cathode and anode half-cells when no electrical current is flowing through the external circuit (open-circuit condition).

It represents the driving force that propels electrons through the external circuit from the anode (oxidation site) to the cathode (reduction site).

Mathematical Formula:

The standard cell electromotive force ($E^\circ_{\text{cell}}$) is calculated using standard reduction potentials:

E°_cell = E°_{cathode (reduction)} - E°_{anode (oxidation)}

Key Rules for Cell EMF:
1. Spontaneous Reaction:    E°_cell > 0 (+ positive value)
2. Non-Spontaneous:        E°_cell < 0 (- negative value)
3. Equilibrium (Dead cell): E°_cell = 0.00 V

Practice Problems

Calculating the EMF of Standard Daniell Cell

Example 1: Calculate the standard electromotive force ($E^\circ_{\text{cell}}$) of a Daniell cell constructed using zinc and copper half-cells given the following standard reduction potentials:
• $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}$
• $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}$

Solution

Identify Cathode and Anode:
The half-cell with the higher (more positive) reduction potential acts as the Cathode (Reduction site).
$\text{Cu}^{2+}/\text{Cu}$ has a higher potential ($+0.34\text{ V}$), so it is the Cathode.
$\text{Zn}^{2+}/\text{Zn}$ has a lower potential ($-0.76\text{ V}$), so it is the Anode.

$$ E°_{cell} = E°_{cathode} - E°_{anode} $$ $$ E°_{cell} = (+0.34 V) - (-0.76 V) $$ $$ E°_{cell} = +0.34 V + 0.76 V $$ $$ E°_{cell} = +1.10 \text{ Volts} $$ The standard cell potential E°_cell is +1.10 V (Spontaneous).

Predicting Reaction Spontaneity

Example 2: Predict whether chlorine gas ($\text{Cl}_2$) will spontaneously oxidize iron(II) ions ($\text{Fe}^{2+}$) to iron(III) ions ($\text{Fe}^{3+}$) in an aqueous solution under standard conditions.
Given:
$E^\circ(\text{Cl}_2/\text{Cl}^-) = +1.36\text{ V}$
$E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}$

Solution

1. Write Half-Cell Reactions:
Chlorine is reduced: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$ (Cathode)
Iron(II) is oxidized: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (Anode)

$$E°_{cell} = E°_\text{cathode} - E°_\text{anode} $$ $$ E°_{cell}= E°(Cl₂/Cl⁻) - E°(Fe³⁺/Fe²⁺) $$ $$ E°_{cell} = (+1.36 V) - (+0.77 V) $$ $$ E°_{cell} = +0.59 \text{ Volts} $$ Since E°_cell (+0.59 V) is positive, the reaction is spontaneous. Chlorine will oxidize Fe²⁺ to Fe³⁺.

Calculating Unknown Electrode Potential

Example 3: An electrochemical cell made of an unknown metal electrode $\text{X}^{2+}/\text{X}$ and a standard silver electrode ($\text{Ag}^+/\text{Ag}$) produces a total cell voltage ($E^\circ_{\text{cell}}$) of $+1.56\text{ V}$. If silver acts as the cathode with $E^\circ(\text{Ag}^+/\text{Ag}) = +0.80\text{ V}$, determine the standard reduction potential of metal $\text{X}$.

Solution

Write out parameters
Cathode ($\text{Ag}^+/\text{Ag}$) = $+0.80\text{ V}$
$E^\circ_{\text{cell}}$ = $+1.56\text{ V}$
Anode ($\text{X}^{2+}/\text{X}$) = $E^\circ_{\text{anode}}$

$$ E°_{cell}= E°_{cathode} - E°_{anode} $$ Algebraic Substitution: $$ +1.56 V = +0.80 V - E°_{anode} $$ $$ E°_{anode} = +0.80 V - 1.56 V $$ $$ E°_{anode} = -0.76 \text{ Volts} $$ The standard reduction potential for metal electrode X²⁺/X is -0.76 V.

Calculating Non-Standard Potential using Nernst Equation

Example 4: Calculate the non-standard electrode potential ($E$) of a copper electrode ($\text{Cu}^{2+}/\text{Cu}$) immersed in a diluted $0.01\text{ M}$ solution of $\text{CuSO}_4$ at $25^\circ\text{C}$.
Given: $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}$, $R = 8.314\text{ J mol}^{-1}\text{K}^{-1}$, $F = 96500\text{ C mol}^{-1}$.

Solution

1. Apply Nernst Equation at $298\text{ K}$ ($25^\circ\text{C}$):
For half-cell reaction: $\text{Cu}^{2+}_{(aq)} + 2e^- \rightarrow \text{Cu}_{(s)}$
number of transferred electrons ($n$) = $2$.

$$ \text{Nernst Equation: }E = E° -\frac{0.0591}{n} × log₁₀\frac{1 }{[Cu²⁺]} $$ $$ Given:$$ $$ E° = +0.34\text{ V} $$ $$ n = 2 $$ $$ [Cu²⁺] = 0.01 M = 10⁻² M $$ $$ E = +0.34 - \frac{0.0591}{2} × log₁₀\frac{1}{0.01} $$ $$ E = +0.34 - (0.02955) × log₁₀(100) $$ $$ E = +0.34 - (0.02955 × 2) $$ $$ E = +0.34 - 0.0591 $$ $$ E = +0.2809\text{ Volts} $$ The non-standard electrode potential E is +0.281 V.

Calculating the Gibbs Free Energy Change ($\Delta G^\circ$) from Cell EMF

Example 5: Calculate the standard Gibbs free energy change ($\Delta G^\circ$) in kilojoules ($\text{kJ}$) for a cell reaction with $E^\circ_{\text{cell}} = +2.71\text{ V}$ involving a two-electron transfer ($n = 2$).
Given: Faraday's constant ($F$) = $96500\text{ C/mol}$.

Solution

1. Formula Relationship:
The thermodynamic link between cell voltage and electrical work is given by: $\Delta G^\circ = -n F E^\circ_{\text{cell}}$

Write out the parameters $$ n = 2 $$ $$ F = 96,500\text{ C/mol} $$ $$ E°_{cell} = +2.71\text{ V (J/C)} $$ $$ ΔG° = - (2) × (96500) × (2.71) $$ $$ ΔG° = - 522,930\text{ J/mol} $$ Convert to kJ: $$ ΔG° = - 522.93 \text{ kJ/mol} $$ The standard Gibbs free energy change ΔG° is -522.93 kJ/mol (Spontaneous).

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Formulas Used:

Standard Cell Potential: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$
Cathode Potential: $E^\circ_{\text{cathode}} = E^\circ_{\text{cell}} + E^\circ_{\text{anode}}$
Anode Potential: $E^\circ_{\text{anode}} = E^\circ_{\text{cathode}} - E^\circ_{\text{cell}}$
Gibbs Free Energy Change: $\Delta G^\circ = -n \cdot F \cdot E^\circ_{\text{cell}}$
Nernst Full Cell Potential (at 298 K): $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log_{10}(Q)$
Nernst Metal Half-Cell Potential (at 298 K): $E = E^\circ - \frac{0.0592}{n} \log_{10}\left(\frac{1}{[M^{n+}]}\right)$
Metal Ion Concentration: $[M^{n+}] = 10^{-\left(\frac{(E^\circ - E) \cdot n}{0.0592}\right)}$
Equilibrium Constant (at 298 K): $K_{\text{eq}} = 10^{\left(\frac{n \cdot E^\circ_{\text{cell}}}{0.0592}\right)}$

Comparison

Property	⚡ High Reduction Potential (+ positive)	🔋 Low Reduction Potential (- negative)
Chemical Behavior	Strong tendency to gain electrons (Reduction)	Strong tendency to lose electrons (Oxidation)
Electrode Role	Acts as the Cathode (+) in a cell	Acts as the Anode (-) in a cell
Agent Strength	Strong Oxidizing Agent (e.g., $\text{F}_2, \text{Cl}_2$)	Strong Reducing Agent (e.g., $\text{Li}, \text{Na}, \text{Zn}$)

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Electrolysis

Electrode Potential ($E$)

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