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Electrochemical Cells (DP IB Chemistry: HL)

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Stewart

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Chemistry Lead

Electrochemical Cells

We have seen previously that redox reactions involve simultaneous oxidation and reduction as electrons flow from the reducing agent to the oxidizing agent
Which way electrons flow depends on the reactivity of the species involved
Redox chemistry has very important applications in electrochemical cells, which come in two types:
- Voltaic cells
- Electrolytic cells

Voltaic cells

A voltaic cell generates a potential difference known as an electromotive force or EMF
The EMF is also called the cell potential and given the symbol E
The absolute value of a cell potential cannot be determined only the difference between one cell and another
- This is analogous to arm-wrestling: you cannot determine the strength of an arm-wrestler unless you compare her to the other competitors

Voltaic (or Galvanic) 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
Each part of the cell is called a half cell
If a rod of metal is dipped into a solution of its own ions, an equilibrium is set up
For example:

Zn (s) ⇌ Zn²⁺(aq) + 2e^–

When a metal is dipped into a solution contains its ions an equilibrium is established between the metal and it ions

This is a half cell and the strip of 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 (eg 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 that a chemical reaction is not taking place – there is simply a potential difference between the rod and the solution

Creating an EMF

If two different electrodes are connected, the potential difference between the two electrodes will cause a current to flow between them. Thus an electromotive force (EMF) is established and the system can generate electrical energy
A typical electrochemical cell can be made by combining a zinc electrode in a solution of zinc sulfate with a copper electrode in a solution of copper sulfate

The zinc-copper voltaic cell (also known as the Daniell Cell)

The circuit must be completed by allowing ions to flow from one solution to the other
This is achieved by means of a salt bridge
- This is often a piece of filter paper saturated with a solution of an inert electrolyte such as KNO₃(aq)

The EMF can be measured using a voltmeter
- Voltmeters have a high resistance so that they do not divert much current from the main circuit
The two half cells are said to be in series as the same current is flowing through both cells
The combination of two electrodes in this way is known as a voltaic cell, and can be used to generate electricity

Cell Potential Calculations

Voltmeters measure potential on the right-hand side of the cell and subtract it from the potential on the left-hand side of the cell

EMF= E_right - E_left

Sometimes this can be hard to remember, but it helps if you remember the phrase 'knives & forks'

E Right minus E left, downloadable IB Chemistry revision notes

You hold your knife in your right hand and your fork in your left hand. EMF is right minus left

If the standard hydrogen electrode is placed on the left-hand side of the voltmeter, then by convention E_leftwill be zero and the EMF of the cell will be the electrode potential of the right-hand electrode
For example, if the standard zinc electrode is connected to the standard hydrogen electrode and the standard hydrogen electrode is placed on the left, the voltmeter measures -0.76V.

Zn²⁺(aq) + 2e^- ⇌ Zn(s)

- The Zn²⁺(aq) + 2e^- ⇌ Zn(s) half-cell thus has an electrode potential of -0.76V
If the Cu²⁺(aq) + 2e^- ⇌ Cu(s) electrode is connected to the standard hydrogen electrode and the standard hydrogen electrode is placed on the left, the voltmeter reads +0.34V
- The Cu²⁺(aq) + 2e^- ⇌ Cu(s) half-cell thus has an electrode potential of +0.34V.

Standard electrode potential

The standard electrode potential of a half-reaction is the emf of a cell where the left-hand electrode is the standard hydrogen electrode and the right-hand electrode is the standard electrode in question
The equation EMF = E_RHS - E_LHS can be applied to electrochemical cells in two ways:
- Calculating an unknown standard electrode potential
- Calculating a cell EMF
To be a standard electrode potential the measurements must be made at standard conditions, namely:
- 1.0 mol dm^-3 ions concentrations
- 100 kPa pressure
- 298 K

Calculating an unknown standard electrode potential

If the RHS and LHS electrode are specified, and the EMF of the cell measured accordingly, then if the E^θ of one electrode is known then the other can be deduced.
- For example, if the standard copper electrode (+0.34 V) is placed on the left, and the standard silver electrode is placed on the right, the EMF of the cell is +0.46 V.
- Calculate the standard electrode potential at the silver electrode.

EMF = E_RHS - E_LHS

+0.46 = E^θ_Ag- (+0.34 V)

E^θ_Ag = 0.46 + 0.34 = +0.80 V

Calculating a cell EMF

If both SEP's are known, the EMF of the cell formed can be calculated if the right-hand electrode and left-hand electrode are specified
- For example, if in a cell the RHS = silver electrode (+0.80V) and LHS is copper electrode (+0.34 V), then

EMF = E_RHS - E_LHS

EMF = +0.80 - 0.34 = +0.46 V

Conventional Representation of Cells

As it is cumbersome and time-consuming to draw out every electrochemical cell in full, a system of notation is used which describes the cell in full, but does not require it to be drawn.
An electrochemical cell can be represented in a shorthand way by a cell diagram (sometimes called cell representations or cell notations)

Conventional cell representation, downloadable IB Chemistry revision notes

The conventional representation of voltaic cells

By convention, the half cell with the greatest negative potential is written on the left of the salt bridge, so E^θ_cell = E^θ_right– E^θ_left
- In this case, E^θ_cell = +0.34 – -0.76 = +1.10 V.
The left cell is being oxidized while the right is being reduced
If there is more than one species in solution, and the species are on different sides of the half-equation, the different species are separated by a comma
This method of representing electrochemical cells is known as the conventional representation of a cell, and it is widely used
If both species in a half reaction are aqueous then an inert platinum electrode is needed which is recorded on the outside of the half cell diagram

Some Examples

For the iron(II) and iron(III) half cell reaction a platinum electrode is needed as an electron carrier
The half equation is

Fe³⁺(aq) + e^- ⇌ Fe²⁺(aq)

So the cell convention as a left hand electrode would be

Pt 丨Fe²⁺(aq), Fe³⁺(aq)

Notice the order must be Fe(II) then Fe(III) as the left side is an oxidation reaction, so Fe(II) is oxidised to Fe(III) by the loss of an electron
The platinum electrode is separated by the phase boundary (vertical solid line), but the iron(II) and iron(III) are separated by a comma since they are in the same phase
Non-metals will also require a platinum electrode
If chlorine is used as an electrode the reduction reaction is

Cl₂(g) + 2e^- ⇌ 2Cl^-(aq)

The conventional representation of the half reaction would be

Cl₂(g), 2Cl^-(aq) | Pt

Notice that the half cell reaction is balanced; however, it would be also correct to write it as

Cl₂(g), Cl^-(aq) | Pt

This is because conventional cell diagrams are not quantitative- they are just representations of the materials and redox processes going on
- Most chemists tend to show them balanced anyway
Combining these two half cells together gives

Pt | Fe²⁺(aq), Fe³⁺(aq) ∥ Cl₂(g), 2Cl^-(aq) | Pt

As you can see the overall cell diagram is not quantitative as the left side is a one electron transfer and the right side is a two electron transfer