Chapter 24 Electrochemistry Flashcards
Standard Cell Potential, ΔE°cell
- The voltage of an electrochemical cell made up of two half-cells at standard conditions.
- Could be any half-cells and neither have to be a standard hydrogen electrode
- Electromotive force (emf)
- Can be measured by voltmeter
Emf of a cell depends on?
- Nature of the electrodes & ions
- Temperature
- Concentrations
Standard Hydrogen Electrode (SHE)
- The ultimate reference when calculating standard cell potential, ΔE°cell
- The SHE is assigned an E° value of 0.00 V
Diagram of SHE
- Hydrogen gas at 100 kPa
- Platinum electrode
- Solution of 1M H+ (aq), e.g. 1M HCl or 0.5M H2SO4
- 298 K or 25 °C
Standard electrode potential
Potential of a half cell compared to the standard hydrogen electrode (SHE)
E°cell formula
E°cell = E°reduction - E°oxidation
- Be careful with sign for E°oxidation
The 4 types of half-cells
- Metal & Ion
- Gases & Ion
- Systems involving oxidation or reduction (ion / ion): solutions of ions in 2 different oxidation states & solutions of oxidising agents in acid solution
Describe the metal / ion half-cell. Use copper (Cu) as an example.
- Reaction
- Electrode
- Solution
- Potential
- R: Cu2+ + 2e- <> Cu (s)
- E: Copper
- S: 1M Cu2+ (aq) / copper sulphate solution
- P: + 0.34 V
Describe the gas / ion half-cell. Use H2 & Cl2 as an example.
- Reaction
- Electrode
- Solution
- Gas
- Potential
H2
- R: 2H+ (aq) + 2e- <> H2 (g)
- E: platinum
- S: H+ (aq) (1M) / 1M HCl / 0.5M H2SO4
- G: H2 at 100 kPa (1 atm)
- P: 0 V
Cl2
- R: Cl2 (g) + 2e- <> 2Cl- (aq)
- E: platinum
- S: Cl- (aq) (1M) / 1M NaCl
- G: Cl2 (g) at 1 atm
- P: +1.36 V
Describe the ion / ion half cell, specifically the solutions of ions in 2 different states. Use Fe3+ & Fe2+ as an example.
- Reaction
- Electrode
- Solution
- Potential
- R: Fe3+ (aq) + e- <> Fe2+ (aq)
- E: Platinum
- S: Fe3+ (aq) (1M) & Fe2+ (aq) (1M)
- P: +0.77 V
Describe the ion / ion half cell, specifically the solutions of oxidising agents in acid solution. Use MnO4- as an example.
- Reaction
- Electrode
- Solution
- Potential
- R: MnO4- (aq) + 8H+ (aq) + 5e- <> Mn2+ (aq) + 4H2O (l)
- E: Platinum
- S: MnO4- (aq) (1M) & Mn2+ (aq) (1M) & H+ (aq)
- P: +1.52 V
Formula for standard cell potential.
Ecellꝋ = Ereductionꝋ – Eoxidationꝋ
- Ecellꝋ > 0 - spontaneous reaction
- Ecellꝋ < 0 - non-spontaneous reaction
Effect of Concentration on E° values.
- Result from changes in concentrations & can be predicted with the Le Chatelier principle.
- Concentration only applies to ions!
What happens when [Cu2+] increases?
- Cu2+ + 2e- <> Cu
- Equilibrium position shifts to the right
- Cu electrode becomes more positively charged
- E° becomes more positive
What happens when [Cu2+] decreases?
- Cu2+ + 2e- <> Cu
- Equilibrium position shifts to the left
- Cu electrode becomes more negatively charged
- E° becomes more negative
Formula for Nernst Equation.
E electrode = E° + (0.059 / z) log [oxidised] / [reduced]
- z = no. of electrons in half-equation
- concentration of solid = 1
- When []s are equimolar, E°cell = E electrode
State the formula that uses the value of E°cell & ΔG° to predict the feasibility of a reaction.
ΔG° = -nE°cellF
- ΔG° - Gibbs free energy in J / mol
- n - no. of moles of electrons transferred in the cell reaction
- F - Faraday constant
- E°cell - standard cell potential
Compare electrochemical & electrolysis cells. (4)
Electrochemical VS Electrolysis:
- Chemical energy –> electrical energy (spontaneous) VS Electrical energy –> chemical energy (non-spontaneous)
- Anode = negative terminal VS Anode = positive terminal
- Cathode = positive terminal VS Cathode = negative terminal
- BOTH: Oxidation happens at anode & Reduction happens at cathode.
Electrolysis
The breaking down of a compound into its elements using an electric current.
Factors affecting selective discharge in electrolysis.
- Position in the redox series - cation with the most positive E° is discharged first (substance that is more easily reduced)
- Relative concentration of ions - ions of higher concentration will be selectively discharged
Order of ease of oxidation at anode.
SO4 2- > NO3 - > Cl- > OH - > Br - > I -
- depends on concentration (refer to slides)
Faraday’s Law.
The amount of substance consumed or produced at one of the electrodes in an electrolytic cell is directly proportional to the amount of electricity that passes through the cell.
Formula for Faraday’s Law.
F = Le
- F = Faraday constant (charge on 1 mole of e-)
- L = Avogadro constant
- e = charge on the electron
What is the charge on 1 electron?
1.60 x 10^-19 C
What is the value of Faraday’s constant?
96500 C
Formula for the quantity of charge passed during electrolysis.
Q = It
- I = current in aWmperes, A
- t = time in seconds, s
- Q = quantity of charge in coulombs, C
What is the value of Avogadro’s constant?
6.02 x 10^23
