Chemistry (9701) | Standard electrode potentials E⦵, standard cell potentials E⦵ cell and the Nernst equation

Electrochemical Cells: $E^\ominus$, $E^\ominus_{\text{cell}}$ and the Nernst Equation

Welcome to one of the most exciting (and sometimes challenging!) topics in Physical Chemistry: Electrochemistry. This chapter is all about understanding how chemical reactions can generate electrical energy, and vice versa. Think of it as the chemistry behind batteries—how do we measure the "chemical push" that makes electrons move?

Don't worry if this seems tricky at first. We will break down the difficult concepts into simple, manageable steps, starting with definitions and building up to the calculations!

Section 1: Standard Electrode Potential ($E^\ominus$)

1.1 Defining the Standard Electrode (Reduction) Potential (24.2.1a)

In electrochemistry, we look at reactions that involve electron transfer (redox reactions). We often study these reactions in two separate compartments called half-cells.

The Standard Electrode Potential ($E^\ominus$) is a measure of the tendency of a species to gain electrons (i.e., undergo reduction).

• $E^\ominus$ is always written for the reduction half-reaction.
• It is measured in Volts (V).
• The value tells you the potential difference relative to a standard reference electrode.

Analogy: Think of $E^\ominus$ as the "greediness" of a chemical species for electrons.

• A large positive $E^\ominus$ (e.g., \(+2.87 \ V\) for $F_2$) means the species loves electrons and is easily reduced. It is a powerful oxidising agent.
• A large negative $E^\ominus$ (e.g., \(-2.37 \ V\) for $Mg^{2+}$) means the species does not like electrons and prefers to be oxidised. It is a powerful reducing agent.

1.2 Standard Conditions (Quick Review)

For the potential to be truly "Standard" ($E^\ominus$), the half-cell must be set up under specific conditions:

• Temperature: \(298 \ K\) (\(25^\circ C\)).
• Concentration: All aqueous ions must have a concentration of \(1.0 \ mol \ dm^{-3}\).
• Pressure: Any gases involved must be at a pressure of \(101 \ kPa\) (1 atmosphere).

1.3 The Standard Hydrogen Electrode (SHE) (24.2.2)

We need a fixed zero point to measure all these potentials against. This reference point is the Standard Hydrogen Electrode (SHE).

The SHE is assigned an electrode potential of exactly \(0.00 \ V\).

How the SHE is constructed:

1. A piece of inert Platinum (Pt) metal (which conducts electrons but doesn't react) is used as the electrode.
2. The Pt electrode is dipped into a solution containing hydrogen ions ($H^+$) at a concentration of \(1.0 \ mol \ dm^{-3}\).
3. Hydrogen gas ($H_2$) is bubbled over the electrode at a pressure of \(101 \ kPa\) and a temperature of \(298 \ K\).

The half-reaction occurring at the SHE is:

\(2H^{+}(aq) + 2e^{-} \rightleftharpoons H_{2}(g) \quad E^\ominus = 0.00 \ V\)

Section 2: Measuring $E^\ominus$ (Half-Cells) (24.2.3)

To measure the $E^\ominus$ of any half-cell, we must connect it to the SHE to form a complete circuit (a cell).

2.1 Setup for Metal/Ion Half-Cells (e.g., Copper) (24.2.3a)

This is the simplest type, consisting of a metal dipping into a solution of its own ions.

• Half-cell setup: A copper strip immersed in \(1.0 \ mol \ dm^{-3}\) $Cu^{2+}(aq)$.
• Connection: Connect the copper half-cell to the SHE using:
  • A voltmeter (to measure the potential difference, $E^\ominus$).
  • A salt bridge (usually filter paper soaked in $KNO_3$), which allows ion movement to complete the circuit but prevents solutions from mixing.
• Measurement: The reading on the voltmeter is the $E^\ominus$ for the copper half-cell (e.g., \(+0.34 \ V\)).
• Half-equation (Reduction): \(Cu^{2+}(aq) + 2e^{-} \rightleftharpoons Cu(s)\)

2.2 Setup for Ion/Ion Half-Cells (e.g., Iron) (24.2.3b)

Sometimes, the half-reaction involves ions of the same element in different oxidation states (e.g., $Fe^{3+}$ and $Fe^{2+}$). Since there is no solid metal, we need an inert electrode.

• Half-cell setup: A Pt electrode immersed in a solution containing both \(1.0 \ mol \ dm^{-3}\) $Fe^{3+}(aq)$ and \(1.0 \ mol \ dm^{-3}\) $Fe^{2+}(aq)$.
• Function of Pt: It acts as a surface for electron transfer, conducting electrons into or out of the solution, but it does not participate in the reaction itself.
• Half-equation (Reduction): \(Fe^{3+}(aq) + e^{-} \rightleftharpoons Fe^{2+}(aq)\)

Did you know? The platinum electrode must be specially coated (platinised) to increase its surface area, ensuring fast equilibrium at the electrode surface.

Section 3: Standard Cell Potential ($E^\ominus_{\text{cell}}$)

3.1 Calculating $E^\ominus_{\text{cell}}$ (24.2.1b, 24.2.4)

A simple cell (battery) is made by combining two different half-cells. The Standard Cell Potential ($E^\ominus_{\text{cell}}$) is the total potential difference (voltage) generated when two half-cells are connected under standard conditions.

We calculate this by finding the difference between the two standard electrode potentials:

\(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}\)

Alternatively, since all $E^\ominus$ values are reduction potentials, we can use:

\(E^\ominus_{\text{cell}} = E^\ominus_{\text{more positive}} - E^\ominus_{\text{less positive}}\)

Step-by-Step Guide to Cell Calculations:

1. Identify the two reduction half-equations and their $E^\ominus$ values.
2. The half-reaction with the more positive $E^\ominus$ will be the reduction reaction (at the cathode).
3. The half-reaction with the less positive $E^\ominus$ will be forced to reverse and become the oxidation reaction (at the anode).
4. Calculate $E^\ominus_{\text{cell}}$ using the formula above.

Example: Zinc/Copper Cell

• \(Cu^{2+} + 2e^{-} \rightleftharpoons Cu \quad E^\ominus = +0.34 \ V\)
• \(Zn^{2+} + 2e^{-} \rightleftharpoons Zn \quad E^\ominus = -0.76 \ V\)

Copper has the more positive $E^\ominus$, so it undergoes reduction. Zinc is oxidised.

\(E^\ominus_{\text{cell}} = (+0.34 \ V) - (-0.76 \ V) = +1.10 \ V\)

3.2 Polarity, Electron Flow, and Feasibility (24.2.5)

The sign of $E^\ominus_{\text{cell}}$ dictates two key aspects:

1. Feasibility (Spontaneity) (24.2.5b):
   If $E^\ominus_{\text{cell}}$ is positive, the reaction is feasible (spontaneous) under standard conditions. (The cell works!)
2. Polarity and Electron Flow (24.2.5a):
   • The Anode (oxidation) is the negative electrode (LHS). Electrons flow away from the anode.
   • The Cathode (reduction) is the positive electrode (RHS). Electrons flow towards the cathode.

   Mnemonic: ANode is NEGative (in a voltaic/galvanic cell, which is spontaneous). RED CAT (Reduction occurs at the Cathode).

3.3 Predicting Reactivity (Oxidising and Reducing Agents) (24.2.6)

We can use $E^\ominus$ values to predict the relative strength of chemicals as redox agents.

• The species with the most positive $E^\ominus$ (on the LHS of its half-equation) is the strongest oxidising agent. It wants to be reduced most strongly.
• The species with the most negative $E^\ominus$ (on the RHS of its half-equation) is the strongest reducing agent. It is most easily oxidised.

General Rule: Any oxidised species (LHS) will spontaneously react with any reduced species (RHS) that sits lower on the $E^\ominus$ list.

Section 4: Constructing Redox Equations (24.2.7)

To get the overall redox equation for a feasible cell, we must combine the two half-equations.

Example: $Zn/Cu$ cell ($E^\ominus_{\text{cell}} = +1.10 \ V$)

1. Reduction (Cathode): \(Cu^{2+}(aq) + 2e^{-} \longrightarrow Cu(s)\)
2. Oxidation (Anode): $Zn$ must be oxidised, so we reverse its half-equation: \(Zn(s) \longrightarrow Zn^{2+}(aq) + 2e^{-}\)
3. Combine: Add the two equations together, ensuring the number of electrons ($e^-$) is balanced and cancelled out. (In this case, 2 electrons cancel.)

Overall Redox Equation: \(Cu^{2+}(aq) + Zn(s) \longrightarrow Cu(s) + Zn^{2+}(aq)\)

Section 5: The Effect of Concentration – The Nernst Equation

The standard potential $E^\ominus$ is only valid under standard conditions (\(1.0 \ mol \ dm^{-3}\)). In a real battery, the concentrations are constantly changing, and therefore the potential ($E$) changes too. This non-standard potential is calculated using the Nernst equation.

5.1 Qualitative Prediction (24.2.8)

Before using the equation, you must be able to predict the direction of the change in potential (E).

Consider a half-reaction: \(\text{Oxidised Species} + z e^{-} \rightleftharpoons \text{Reduced Species}\)

• Increase Concentration of Oxidised Species (Reactant): According to Le Chatelier’s Principle, the equilibrium shifts to the right (to favour reduction). This makes the potential more positive (easier to reduce).
• Increase Concentration of Reduced Species (Product): The equilibrium shifts to the left (to favour oxidation). This makes the potential less positive (more negative).

5.2 The Nernst Equation (24.2.9)

The Nernst equation allows us to calculate the actual potential ($E$) when concentrations are non-standard. The simplified form used in A-Level Chemistry is:

$$E = E^\ominus + \frac{0.059}{z} \log \frac{[\text{oxidised species}]}{[\text{reduced species}]}$$

Where:

• \(E\) is the non-standard electrode potential (V).
• \(E^\ominus\) is the standard electrode potential (V).
• \(z\) is the number of electrons transferred in the half-reaction.
• \([\text{oxidised species}]\) and \([\text{reduced species}]\) are the non-standard concentrations of the aqueous ions involved.
• Note: Solids (like $Cu(s)$) are omitted from the concentration ratio because their effective concentration is constant (usually taken as 1).

Application Example 1: Copper Half-Cell

The reduction equation is: \(Cu^{2+}(aq) + 2e^{-} \rightleftharpoons Cu(s)\). Here, \(z=2\).

The Nernst equation becomes:

$$E = E^\ominus + \frac{0.059}{2} \log [Cu^{2+}(aq)]$$

(Since the reduced species is the solid $Cu$, the term in the denominator is 1 and can be ignored.)

Application Example 2: Iron Half-Cell

The reduction equation is: \(Fe^{3+}(aq) + e^{-} \rightleftharpoons Fe^{2+}(aq)\). Here, \(z=1\).

The Nernst equation becomes:

$$E = E^\ominus + \frac{0.059}{1} \log \frac{[Fe^{3+}(aq)]}{[Fe^{2+}(aq)]}$$

Note how both ions are aqueous, so both concentrations must be included in the concentration ratio.

Section 6: Relating Cell Potential to Gibbs Free Energy ($\Delta G^\ominus$) (24.2.10)

In Chemical Energetics, we learned that the feasibility of a reaction is determined by the sign of the Gibbs Free Energy change ($\Delta G^\ominus$). Electrochemistry provides a direct link between this thermodynamic concept and the measured potential ($E^\ominus_{\text{cell}}$).

The electrical energy generated by a cell is equal to the decrease in Gibbs Free Energy.

$$ \Delta G^\ominus = -nE^\ominus_{\text{cell}}F $$

Where:

• \(\Delta G^\ominus\) is the standard Gibbs free energy change (\(J \ mol^{-1}\) or \(kJ \ mol^{-1}\)).
• \(n\) is the amount of charge transferred (in moles of electrons) in the overall reaction. (You find this by balancing the half-equations).
• \(E^\ominus_{\text{cell}}\) is the standard cell potential (V).
• \(F\) is the Faraday constant (\(96,500 \ C \ mol^{-1}\)). This is the charge carried by one mole of electrons.

Feasibility Check:

Since the Faraday constant ($F$) and moles of electrons ($n$) are always positive:

• If \(E^\ominus_{\text{cell}}\) is positive, then \(\Delta G^\ominus\) must be negative. This means the reaction is spontaneous (feasible).
• If \(E^\ominus_{\text{cell}}\) is negative, then \(\Delta G^\ominus\) must be positive. This means the reaction is non-spontaneous.

This equation officially confirms that a positive cell voltage guarantees a reaction will occur!