Standard electrode potentials E ⦵ ; standard cell potentials E ⦵ cell and the Nernst equation Flashcards
standard electrode (reduction) potential
a value which shows how easily a substance is reduced (gains electrons)
redox equilibrium
if you dipped a metal rod into a solution which contained metal ions, there would be metal atoms losing electrons to form metal ions and at the same time, metal ions gaining electrons to become metal atoms
electrons are written
on the left-hand side (demonstrating reduction)
why different species will have different electrode (reduction) potentials
position of equilibrium is different for different species
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
more negative (or less positive) an electrode potential
-the less likely it is that reduction of that species will occur
-The equilibrium position lies more to the left
factors that effect electrode potential
-Temperature
-Pressure of gases
-Concentration of reagents
What are the standard conditions that have to be used when comparing electrode potentials
-Ion concentration of 1.00 mol dm-3
-A temperature of 298 K
-A pressure of 1 atm
electrode potentials are measured relative to
standard hydrogen electrode
standard hydrogen electrode is given a value of
0.00 V
standard electrode potential (Eꝋ)
the voltage produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions
standard cell potential (Ecellꝋ)
the difference in Eꝋ between two half-cells
standard hydrogen electrode
-is a half-cell used as reference electrodes
-consists of:
1)Hydrogen gas in equilibrium with H+ ions of concentration 1.00 mol dm-3 (at 1 atm)
2)An inert platinum electrode that is in contact with the hydrogen gas and H+ ions
method used to measure the standard electrode potentials of metal/metal ion half-cell
-An example of a metal/metal ion half-cell is the Ag+/ Ag half-cell
-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
method used to measure the standard electrode potentials of non-metal/non-metal ion half-cell
-platinum wire or foil is used as an electrode to make electrical contact with the solution
-An example of a non-metal/non-metal ion is the Br2/Br- half-cell
-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
-The Ecellꝋ is: 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
method used to measure the standard electrode potentials of ions of the same element in different oxidation states
-platinum electrode is 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) Eꝋ = +1.52 V
2H+ (aq) + 2e- ⇌ H2 (g) Eꝋ = 0.00 V
-The MnO4-/Mn2+ - half-cell is the positive pole and the H+/H2 is the negative pole
-The Ecellꝋ = (+ 1.52) - (0.00) = + 1.52 V
the electrode with the more positive Eꝋ value is the
positive pole
Reduction occurs at the
positive pole
Oxidation occurs at the
negative pole
Ecellꝋ =
Eꝋ of best reducing agent in solution - Eꝋ of best oxidising agent in solution
direction of electron flow in electrochemical cell
flow of electrons is from the negative pole to the positive pole
The more positive the Eꝋ 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 Eꝋ 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 when
the Ecellꝋ is positive
Constructing redox equations
Step 1: Determine in which half-cell the oxidation and in which half-cell the reduction reaction takes place
Step 2: Write down the half equations for each half-cell
Step 3: Balance the number of electrons in both half-equations
Step 4 - Add up the two half-equations
Step 5 - Cancel out the electrons (and H+ ions and H2O molecules if any present) to find the overall redox reaction
Faraday’s constant can be used to calculate
-The mass of a substance deposited at an electrode
-The volume of gas liberated at an electrode
To calculate the mass of a substance deposited at an electrode
-Write the half-equation at the electrode
-Determine the number of coulombs needed to form one mole of substance at the specific electrode using Faraday’s constant
-Calculate the charge transferred during electrolysis
-Use simple proportion and the relative atomic mass of the substance to find its mass
To calculate volume of gas liberated at an electrode
-Write the half-equation at the electrode
-Determine the number of coulombs needed to form one mole of substance at the specific electrode using Faraday’s constant
-Calculate the charge transferred during electrolysis
-Use simple proportion and the relationship 1 mol of gas occupies 24.0 dm3 at room temperature
The Eꝋ values of a species indicate how easily they can get
oxidised or reduced
More positive (less negative) Eꝋ values indicate that
-The species is easily reduced
-The species is a better oxidising agent
Less positive (more negative) Eꝋ values indicate that
-The species is easily oxidised
-The species is a better reducing agent
If the concentration of the species on the left is increased
-the position of equilibrium will shift to the right
-This means that the species on the left gets more easily reduced
-The E value becomes more positive (or less negative)
If the concentration of the species on the right is increased
-the position of equilibrium will shift to the left
-This means that the species on the left gets less easily reduced
-The E value becomes less positive (or more negative)
The effect of changes in temperature and ion concentration on the Ecell can be deduced using the
Nernst equation
E =
E⦵+ (0.059/z) log [oxidised species]/[reduced species]
or
E⦵+RT/zF ln[oxidised species]/[reduced species]
ΔGꝋ =
n x Ecellꝋ x F
n = number of electrons transferred in the reaction
Ecellꝋ = standard cell potential (V)
F = Faraday constant (96 500 C mol-1)
How to calculate standard free energy change using the standard cell potential of an electrochemical cell
Step 1: Determine the two half-equations and their Eꝋ using the Data booklet
Step 2 : Calculate the Ecellꝋ
Step 3: Determine the number of electrons transferred in the reaction
Step 4: Substitute the values in for the standard Gibbs free energy equation
