Electrode Potentials & Cells (Oxford AQA International A Level Chemistry)

Cell Representation

• Electrochemical cells generate electricity from spontaneous redox reactions

• For example:

Zn (s) + CuSO₄ (aq) → Cu (s) + ZnSO₄ (aq)

• Instead of electrons being transferred directly from the zinc to the copper ions, a cell is built which separates the two redox processes

• For example:

Zn (s) ⇌ Zn²⁺ (aq) + 2e^– 

• If a rod of metal is dipped into a solution of its own ions, an equilibrium is set up

  • Each part of the cell is called a half cell

  • The metal is an electrode

• The position of the equilibrium determines the potential difference between the metal strip and the solution of metal

• The Zn atoms on the rod can deposit two electrons on the rod and move into solution as Zn²⁺ ions:

Zn (s) ⇌ Zn²⁺ (aq) + 2e^– 

• This process would result in an accumulation of negative charge on the zinc rod

• Alternatively, the Zn²⁺ ions in solution could accept two electrons from the rod and move onto the rod to become Zn atoms:

Zn²⁺ (aq) + 2e^– ⇌ Zn (s)

• This process would result in an accumulation of positive charge on the zinc rod

• In both cases, a potential difference is set up between the rod and the solution

  • This is known as an electrode potential

• A similar electrode potential is set up if a copper rod is immersed in a solution containing copper ions (e.g. CuSO₄), due to the following processes:

Cu²⁺ (aq) + 2e^– ⇌ Cu (s)   – reduction (rod becomes positive)

Cu (s) ⇌ Cu²⁺ (aq) + 2e^–    – oxidation (rod becomes negative)

• NOTE: A chemical reaction is not taking place

  • It is simply a potential difference between the rod and the solution

• The potential difference depends on the:

  • Nature of the ions in solution

  • Concentration of the ions in solution

  • Type of electrode used

  • Temperature

Electrode potential

• The electrode (reduction) potential (E) is a value which shows how easily a substance is reduced

• These are demonstrated using reversible, ionic half-equations

  • This is because there is a redox equilibrium between two related species that are in different oxidation states

• When writing half-equations for this topic, the electrons will always be written on the left-hand side (demonstrating reduction)

• The position of equilibrium is different for different species

  • This is why different species will have different electrode (reduction) potentials

• The more positive (or less negative) an electrode potential, the more likely it is for that species to undergo reduction

  • The equilibrium position lies more to the right

• For example, the positive electrode potential of bromine below, suggests that it is likely to get reduced and form bromide (Br^-) ions

Br₂ (l) + 2e^- ⇌ 2Br^- (aq)        E = +1.09 V

• The more negative (or less positive) the electrode potential, the less likely it is that reduction of that species will occur

  • The equilibrium position lies more to the left

• For example, the negative electrode potential of sodium suggests that it is unlikely that the sodium (Na⁺) ions will be reduced to sodium (Na) atoms

Na⁺ (aq) + e^- ⇌ Na (s)        E = -2.71 V

Conventional Representation of Cells

• Chemists use a type of shorthand convention to represent electrochemical cells

• In this convention:

  • A solid vertical (or slanted) line shows a phase boundary, that is an interface between a solid and a solution

  • A double vertical line (sometimes shown as dashed vertical lines) represents a salt bridge

    • A salt bridge has mobile ions that complete the circuit

    • Potassium chloride and potassium nitrate are commonly used to make the salt bridge as chlorides and nitrates are usually soluble

    • This should ensure that no precipitates form which can affect the equilibrium position of the half cells

  • The substance with the highest oxidation state in each half cell is drawn next to the salt bridge

  • The cell potential difference is shown with the polarity of the right hand electrode

• The cell convention for the zinc and copper cell would be

Zn (s) ∣ Zn²⁺ (aq) ∥ Cu²⁺ (aq) ∣ Cu (s)                  E _cell = +1.10 V

• This tells us the copper half cell is more positive than the zinc half cell, so that electrons would flow from the zinc to the copper

• The same cell can be written as:

Cu (s) ∣ Cu²⁺ (aq) ∥ Zn²⁺ (aq) ∣ Zn (s)                  E _cell = -1.10 V

• The polarity of the right hand half cell is negative, so we can still tell that electrons flow from the zinc to the copper half cell

Standard Hydrogen Electrode

• The standard hydrogen electrode is a half-cell used as a reference electrode and consists of:

  • An inert platinum electrode that is in contact with the hydrogen gas and H⁺ ions

  • Hydrogen gas in equilibrium with H⁺ ions of concentration 1.00 mol dm^-3 (at 100 kPa)

2H⁺ (aq) + 2e^- ⇌ H₂ (g)

• When the standard hydrogen electrode is connected to another half-cell, the standard electrode potential of that half-cell can be read off a high resistance voltmeter

• There are three different types of half-cells that can be connected to a standard hydrogen electrode

  • A metal / metal ion half-cell

  • A non-metal / non-metal ion half-cell

  • An ion / ion half-cell (the ions are in different oxidation states)

Metal / metal ion half-cell

• An example of a metal / metal ion half-cell is the Ag⁺ / Ag half-cell

  • Ag is the metal

  • Ag⁺ is the metal ion

• This half-cell is connected to a standard hydrogen electrode and the two half-equations are:

Ag⁺ (aq) + e^- ⇌ Ag (s)        E^ꝋ = + 0.80 V

2H⁺ (aq) + 2e^- ⇌ H₂ (g)        E^ꝋ = 0.00 V 

• The Ag⁺/ Ag half-cell has a more positive E^ꝋ value

  • Therefore, it is the positive pole

  • So, the H⁺ / H₂ half-cell is the negative pole

• The standard cell potential (E_cell^ꝋ) is:

E_cell^ꝋ = (+ 0.80) - (0.00) = + 0.80 V

• The Ag⁺ ions are more likely to get reduced than the H⁺ ions as it has a greater E^ꝋ value

  • Reduction occurs at the positive electrode

  • Oxidation occurs at the negative electrode

Non-metal / non-metal ion half-cell

• In a non-metal / non-metal ion half-cell, platinum wire or foil is used as an electrode to make electrical contact with the solution

  • Like graphite, platinum is inert and does not take part in the reaction

  • The redox equilibrium is established on the platinum surface

• An example of a non-metal / non-metal ion is the Br₂ / Br^- half-cell

  • Br₂ is the non-metal

  • Br^- is the non-metal ion

• The half-cell is connected to a standard hydrogen electrode and the two half-equations are:

Br₂ (aq) + 2e^- ⇌ 2Br^- (aq)        E^ꝋ = +1.09 V

2H⁺ (aq) + 2e^- ⇌ H₂ (g)        E^ꝋ = 0.00 V

• The Br₂ / Br^- half-cell has a more positive E^ꝋ value

  • Therefore, it is the positive pole

  • So, the H⁺ / H₂ half-cell is the negative pole

• The standard cell potential (E_cell^ꝋ) is:

E_cell^ꝋ = (+ 1.09) - (0.00) = + 1.09 V

• The Br₂ molecules are more likely to get reduced than H⁺ as they have a greater E^ꝋ value

Ion / Ion half-cell

• A platinum electrode is again used to form a half-cell of ions that are in different oxidation states

• An example of such a half-cell is the MnO₄^- / Mn²⁺ half-cell

  • MnO₄^- is an ion containing Mn with oxidation state +7

  • The Mn²⁺ ion contains Mn with oxidation state +2

• This half-cell is connected to a standard hydrogen electrode and the two half-equations are:

MnO₄^- (aq) + 8H⁺ (aq) + 5e^- ⇌ Mn²⁺ (aq) + 4H₂O (l)       E^ꝋ = +1.52 V

2H⁺ (aq) + 2e^- ⇌ H₂ (g)       E^ꝋ = 0.00 V   

• The H⁺ ions are also present in the half-cell as they are required to convert MnO₄^- into Mn²⁺ ions

• The MnO₄^- / Mn²⁺ half-cell has a more positive E^ꝋ value

  • Therefore, it is the positive pole

  • So, the H⁺ / H₂ half-cell is the negative pole

• The standard cell potential (E_cell^ꝋ) is:

E_cell^ꝋ = (+ 1.52) - (0.00) = + 1.52 V

• The MnO₄^- ions are more likely to get reduced than H⁺ as they have a greater E^ꝋ value

Standard Electrode Potentials

• The position of equilibrium and the electrode potential depends on factors such as:

  • Temperature

  • Pressure of gases

  • Concentration of reagents

• To compare the electrode potentials of different species, they have to be measured against a common reference or standard

• Standard conditions have to be used when comparing electrode potentials

• Standard conditions are:

  • An ion concentration of 1.00 mol dm^-3

  • A temperature of 298 K

  • A pressure of 100 kPa

• Standard measurements are made using a high resistance voltmeter

  • This is so no current flows and the maximum potential difference is achieved

• Electrode potentials are measured relative to a standard hydrogen electrode

  • The standard hydrogen electrode is given a value of 0.00 V

  • All other electrode potentials are compared to this standard

• This means that the electrode potentials are always referred to as a standard electrode potential

• The standard electrode potential is the potential difference produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions

  • Standard electrode potential is given the symbol E^ꝋ

The electrochemical series

• The E^ꝋ values of a species indicate how easily they can get oxidised or reduced

  • The values indicate the relative reactivity of elements, compounds and ions as oxidising agents or reducing agents

• The electrochemical series is a list of various redox equilibria in order of decreasing E^ꝋ values

• More positive (less negative) E^ꝋ values indicate that:

  • The species is easily reduced

  • The species is a better oxidising agent

• Less positive (more negative) E^ꝋ values indicate that:

  • The species is easily oxidised

  • The species is a better reducing agent

• For example, the standard electrode potential of bromine suggests that relative to the hydrogen half-cell it is more likely to get reduced, as it has a more positive E^ꝋ value

Br₂ (l) + 2e^– ⇌ 2Br^– (aq)        E^ꝋ = +1.09 V          

2H⁺ (aq) + 2e^– ⇌ H₂ (g)        E^ꝋ = 0.00 V

Predicting redox reactions

• The electrochemical series can also be used to predict:

  • The direction of electron flow

  • The feasibility of a redox reaction

Electron flow

• The direction of electron flow can be determined by comparing the E^ꝋ values of two half-cells in an electrochemical cell

Cl₂ (g) + 2e^- ⇌ 2Cl^- (aq)        E^ꝋ = +1.36 V

Cu²⁺ (aq) + 2e^- ⇌ Cu (s)        E^ꝋ = +0.34 V

• The Cl₂ / Cl^- half-cell has a more positive E^ꝋ value

  • So, it is the positive pole

  • The Cl₂ will more readily accept electrons from the Cu²⁺ / Cu half-cell

  • The Cl₂ gets more readily reduced

• The electrons flow from the Cu²⁺ / Cu half-cell to the Cl₂ / Cl^- half-cell

  • The flow of electrons is from the negative pole to the positive pole

Reaction feasibility

• The more positive the E^ꝋ value, the easier it is to reduce the species on the left of the half-equation

  • The reaction will tend to proceed in the forward direction

• The less positive the E^ꝋ value, the easier it is to oxidise the species on the right of the half-equation

  • The reaction will tend to proceed in the backward direction

• For example, two half-cells in the following electrochemical cell are:

Cl₂ (g) + 2e^- ⇌ 2Cl^- (aq)        E^ꝋ = +1.36 V

Cu²⁺ (aq) + 2e^- ⇌ Cu (s)        E^ꝋ = +0.34 V

• Cl₂ molecules are reduced as they have a more positive E^ꝋ value

• The chemical reaction that occurs in this half cell is:

Cl₂ (g) + 2e^- ⇌ 2Cl^- (aq)          

• Cu²⁺ ions are oxidised as they have a less positive E^ꝋ value

• The chemical reaction that occurs in this half cell is:

Cu (s) ⇌ Cu²⁺ (aq) + 2e^-

• Combining both equations gives:

Cu (s) + Cl₂ (g) + 2e^- ⇌ 2Cl^- (aq) + Cu²⁺ (aq) + 2e^-

• Cancelling out the electrons on both sides gives the overall equation:

Cu (s) + Cl₂ (g) ⇌ 2Cl^- (aq) + Cu²⁺ (aq)

OR

Cu (s) + Cl₂ (g) ⇌ CuCl₂ (s)

• A reaction is feasible or spontaneous when the standard cell potential, E_cell^ꝋ, is positive

• To calculate standard cell potential:

E_cell^ꝋ = E_reduction^ꝋ - E_oxidation^ꝋ

• Feasibility of the forward reaction:

  • Cu atoms are oxidised to become Cu²⁺ ions

  • Cl₂ molecules are reduced to become Cl^- ions

  • E_cell^ꝋ = E_reduction^ꝋ - E_oxidation^ꝋ

    • E_cell^ꝋ = (+1.36) - (+0.34)

    • E_cell^ꝋ = +1.02 V

  • E_cell^ꝋ is a positive value so the forward reaction is feasible

• Feasibility of the backward reaction:

  • Cu²⁺ ions are reduced to become Cu atoms

  • Cl^- ions are oxidised to become Cl₂ molecules

  • E_cell^ꝋ = E_reduction^ꝋ - E_oxidation^ꝋ

    • E_cell^ꝋ = (+0.34) - (+1.36)

    • E_cell^ꝋ = -1.02 V

  • E_cell^ꝋ is a negative value so the backward reaction is not feasible