Standard electrode potentials E ⦵ ; standard cell potentials E ⦵ cell and the Nernst equation

Question 1

standard electrode (reduction) potential

Answer 1

a value which shows how easily a substance is reduced (gains electrons)

Question 2

redox equilibrium

Answer 2

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

Question 3

electrons are written

Answer 3

on the left-hand side (demonstrating reduction)

Question 4

why different species will have different electrode (reduction) potentials

Answer 4

position of equilibrium is different for different species

Question 5

more positive (or less negative) an electrode potential

Answer 5

-the more likely it is for that species to undergo reduction
-The equilibrium position lies more to the right

Question 6

more negative (or less positive) an electrode potential

Answer 6

-the less likely it is that reduction of that species will occur
-The equilibrium position lies more to the left

Question 7

factors that effect electrode potential

Answer 7

-Temperature
-Pressure of gases
-Concentration of reagents

Question 8

What are the standard conditions that have to be used when comparing electrode potentials

Answer 8

-Ion concentration of 1.00 mol dm-3
-A temperature of 298 K
-A pressure of 1 atm

Question 9

electrode potentials are measured relative to

Answer 9

standard hydrogen electrode

Question 10

standard hydrogen electrode is given a value of

Answer 10

0.00 V

Question 11

standard electrode potential (Eꝋ)

Answer 11

the voltage produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions

Question 12

standard cell potential (Ecellꝋ)

Answer 12

the difference in Eꝋ between two half-cells

Question 13

standard hydrogen electrode

Answer 13

-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

Question 14

method used to measure the standard electrode potentials of metal/metal ion half-cell

Answer 14

-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

Question 15

method used to measure the standard electrode potentials of non-metal/non-metal ion half-cell

Answer 15

-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

Question 16

method used to measure the standard electrode potentials of ions of the same element in different oxidation states

Answer 16

-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

Question 17

the electrode with the more positive Eꝋ value is the

Answer 17

positive pole

Question 18

Reduction occurs at the

Answer 18

positive pole

Question 19

Oxidation occurs at the

Answer 19

negative pole

Question 20

Ecellꝋ =

Answer 20

Eꝋ of best reducing agent in solution - Eꝋ of best oxidising agent in solution

Question 21

direction of electron flow in electrochemical cell

Answer 21

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

Question 22

The more positive the Eꝋ value

Answer 22

-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

Question 23

The less positive the Eꝋ value

Answer 23

-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

Question 24

A reaction is feasible when

Answer 24

the Ecellꝋ is positive

Question 25

Constructing redox equations

Answer 25

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

Question 26

Faraday’s constant can be used to calculate

Answer 26

-The mass of a substance deposited at an electrode
-The volume of gas liberated at an electrode

Question 27

To calculate the mass of a substance deposited at an electrode

Answer 27

-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

Question 28

To calculate volume of gas liberated at an electrode

Answer 28

-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

Question 29

The Eꝋ values of a species indicate how easily they can get

Answer 29

oxidised or reduced

Question 30

More positive (less negative) Eꝋ values indicate that

Answer 30

-The species is easily reduced
-The species is a better oxidising agent

Question 31

Less positive (more negative) Eꝋ values indicate that

Answer 31

-The species is easily oxidised
-The species is a better reducing agent

Question 32

If the concentration of the species on the left is increased

Answer 32

-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)

Question 33

If the concentration of the species on the right is increased

Answer 33

-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)

Question 34

The effect of changes in temperature and ion concentration on the Ecell can be deduced using the

Answer 34

Nernst equation

Question 35

E =

Answer 35

E⦵+ (0.059/z) log [oxidised species]/[reduced species]
or
E⦵+RT/zF ln[oxidised species]/[reduced species]

Question 36

ΔGꝋ =

Answer 36

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)

Question 37

How to calculate standard free energy change using the standard cell potential of an electrochemical cell

Answer 37

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