Measuring the Standard Electrode Potential (Cambridge (CIE) A Level Chemistry): Revision Note
Measuring the Standard Electrode Potential
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)
When a half-cell is connected to a standard hydrogen electrode, or when two half-cells are connected, a salt bridge is required
A salt bridge has mobile ions that complete the circuit
A salt bridge is typically a strip of filter paper soaked in a saturated solution of potassium nitrate or potassium chloride as nitrates and chlorides are usually soluble
This should ensure that no precipitates form which can affect the equilibrium position of the half-cells
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) Eꝋ = + 0.80 V
2H+ (aq) + 2e- ⇌ H2 (g) Eꝋ = 0.00 V
Since the Ag+ / Ag half-cell has a more positive Eꝋ value, this is the positive pole and the H+ / H2 half-cell is the negative pole
The standard cell potential (Ecellꝋ) is Ecellꝋ = (+ 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 Eꝋ value
Reduction occurs at the positive pole
Oxidation occurs at the negative pole
Example of an electrochemical cell containing a metal / metal ion half-cell
Under standard conditions, a metal / metal ion half-cell is connected to the standard hydrogen electrode to measure the cell potential
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
Br 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 (l) + 2e- ⇌ 2Br- (aq) Eꝋ = +1.09 V
2H+ (aq) + 2e- ⇌ H2 (g) Eꝋ = 0.00 V
The Br2 / Br- half-cell is the positive pole and the H+ / H2 is the negative pole
Ecellꝋ = (+ 1.09) - (0.00) = + 1.09 V
The Br2 molecules are more likely to get reduced than H+ as they have a greater Eꝋ value
Example of an electrochemical cell containing a non-metal / non-metal ion half-cell
Under standard conditions, a non-metal / non-metal ion half-cell is connected to the standard hydrogen electrode to measure the cell potential
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 an oxidation number of +7
The Mn2+ ion contains Mn with an oxidation number of +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) Eꝋ = +1.52 V
2H+ (aq) + 2e- ⇌ H2 (g) Eꝋ = 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 is the positive pole and the H+ / H2 is the negative pole
Ecellꝋ = (+ 1.52) - (0.00) = + 1.52 V
Example of an electrochemical cell containing an ion / ion half-cell
Under standard conditions, an ion / ion half-cell is connected to the standard hydrogen electrode to measure the cell potential
Standard Cell Potential: Direction of Electron Flow & Feasibility
Direction of electron flow
The direction of electron flow can be determined by comparing the Eꝋ values of two half-cells in an electrochemical cell
2Cl2 (g) + 2e- ⇌ 2Cl- (aq) Eꝋ = +1.36 V
Cu2+ (aq) + 2e- ⇌ Cu (s) Eꝋ = +0.34 V
The Cl2 more readily accept electrons from the Cu2+/Cu half-cell
This is the positive pole
Cl2 gets more readily reduced
The Cu2+ more readily loses electrons to the Cl2/Cl- half-cell
This is the negative pole
Cu2+ gets more readily oxidised
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
Flow of electrons through an electrochemical cell
The electrons flow through the wires from the negative pole to the positive pole
Feasibility
The Eꝋ values of a species indicate how easily they can get oxidised or reduced
The more positive the 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 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
A reaction is feasible (likely to occur) when the Ecellꝋ is positive
For example, two half-cells in the following electrochemical cell are:
Cl2 (g) + 2e- ⇌ 2Cl- (aq) Eꝋ = +1.36 V
Cu2+ (aq) + 2e- ⇌ Cu (s) Eꝋ = +0.34 V
Cl2 molecules are reduced as they have a more positive Eꝋ 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 Eꝋ value
The chemical reaction that occurs in this half-cell is:
Cu (s) → Cu2+ (aq) + 2e-
The overall equation of the electrochemical cell is (after cancelling out the electrons):
Cu (s) + Cl2 (g) → 2Cl- (aq) + Cu2+ (aq)
OR
Cu (s) + Cl2 (g) → CuCl2 (s)
The forward reaction is feasible (spontaneous) as it has a positive Eꝋ value of +1.02 V ((+1.36) - (+0.34))
The backward reaction is not feasible (not spontaneous) as it has a negative Eꝋ value of -1.02 ((+0.34) - (+1.36))
Reaction feasibility and the standard cell potential
A reaction is feasible when the standard cell potential Eꝋ is positive
Examiner Tips and Tricks
Remember that the electrons only move through the wires in the external circuit and not through the electrolyte solution.
Redox Equations
The main ways to construct redox equations using the relevant half-equations:
Using changes in oxidation numbers to help balance chemical equations
Using the number of electrons for each half-cell
Both of these are discussed in the Redox Reactions topic
Interpreting the information given to you and predicting any other chemicals involved in the reaction
Worked Example
Write the balanced chemical equation for hydrogen iodide reacting with sulfuric acid to form hydrogen sulfide, iodine and one other product
Answer:
There are 2 possible methods:
Using ionic half-equations
Interpreting the information and predicting any other chemicals involved
Using ionic half-equations method:
Step 1: Identify possible half-equations from the information
Iodide → iodine
Sulfuric acid → hydrogen sulfide
Step 2: Construct the half equations
Iodide → iodine
I– → I2
This requires 2 iodide ions and 2 electrons as products to balance the equation
2I– → I2 + 2e–
Sulfuric acid → hydrogen sulfide
H2SO4 → H2S
The 4 oxygen atoms will form 4 water molecules as products
H2SO4 → H2S + 4H2O
The 8 hydrogen atoms in the water molecules will require 8 protons as reactants
H2SO4 + 8H+ → H2S + 4H2O
The 8 protons will require 8 electrons to balance the charge
H2SO4 + 8H+ + 8e– → H2S + 4H2O
Step 3: Combine the half-equations
Iodide → iodine
2I– → I2 + 2e–
The iodide half-equation needs to be multiplied by 4 to have the same number of electrons as the sulfuric acid half-equation
8I– → 4I2 + 8e–
Sulfuric acid → hydrogen sulfide
H2SO4 + 8H+ + 8e– → H2S + 4H2O
Combining the half-equations
8I– + H2SO4 + 8H+ + 8e– → 4I2 + 8e– H2S + 4H2O
Cancelling, where appropriate
The electrons cancel on both sides
The 8I– and 8H+ can be re-written as 8HI
8HI + H2SO4 → 4I2 + H2S + 4H2O
Interpreting and predicting method:
Step 1: Start with what you know:
HI + H2SO4 → H2S + I2
Step 2: Consider any elements that are not accounted for
The only element that is not currently considered is oxygen
Step 3: Make a common and appropriate suggestion for the missing product
Most of these questions are in solution so there is always H2O, H+ and OH- available
Missing product suggestion = 4H2O
HI + H2SO4 → H2S + I2 + 4H2O
Step 4: Balance the remaining chemicals
8HI + H2SO4 → H2S + 4I2 + 4H2O
Examiner Tips and Tricks
Similar approaches can be used to balance more complicated ionic half-equations
In these situations, you will have H2O, H+, OH- and electrons available
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