Chapter 24 Electrochemistry Flashcards

1
Q

Standard Cell Potential, ΔE°cell

A
  • 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
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2
Q

Emf of a cell depends on?

A
  • Nature of the electrodes & ions
  • Temperature
  • Concentrations
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3
Q

Standard Hydrogen Electrode (SHE)

A
  • The ultimate reference when calculating standard cell potential, ΔE°cell
  • The SHE is assigned an E° value of 0.00 V
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4
Q

Diagram of SHE

A
  • 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
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5
Q

Standard electrode potential

A

Potential of a half cell compared to the standard hydrogen electrode (SHE)

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6
Q

E°cell formula

A

E°cell = E°reduction - E°oxidation

  • Be careful with sign for E°oxidation
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7
Q

The 4 types of half-cells

A
  • 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
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8
Q

Describe the metal / ion half-cell. Use copper (Cu) as an example.
- Reaction
- Electrode
- Solution
- Potential

A
  • R: Cu2+ + 2e- <> Cu (s)
  • E: Copper
  • S: 1M Cu2+ (aq) / copper sulphate solution
  • P: + 0.34 V
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9
Q

Describe the gas / ion half-cell. Use H2 & Cl2 as an example.
- Reaction
- Electrode
- Solution
- Gas
- Potential

A

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

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10
Q

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

A
  • R: Fe3+ (aq) + e- <> Fe2+ (aq)
  • E: Platinum
  • S: Fe3+ (aq) (1M) & Fe2+ (aq) (1M)
  • P: +0.77 V
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11
Q

Describe the ion / ion half cell, specifically the solutions of oxidising agents in acid solution. Use MnO4- as an example.
- Reaction
- Electrode
- Solution
- Potential

A
  • 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
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12
Q

Formula for standard cell potential.

A

Ecellꝋ = Ereductionꝋ – Eoxidationꝋ
- Ecellꝋ > 0 - spontaneous reaction
- Ecellꝋ < 0 - non-spontaneous reaction

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13
Q

Effect of Concentration on E° values.

A
  • Result from changes in concentrations & can be predicted with the Le Chatelier principle.
  • Concentration only applies to ions!
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14
Q

What happens when [Cu2+] increases?

A
  • Cu2+ + 2e- <> Cu
  • Equilibrium position shifts to the right
  • Cu electrode becomes more positively charged
  • E° becomes more positive
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15
Q

What happens when [Cu2+] decreases?

A
  • Cu2+ + 2e- <> Cu
  • Equilibrium position shifts to the left
  • Cu electrode becomes more negatively charged
  • E° becomes more negative
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16
Q

Formula for Nernst Equation.

A

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

17
Q

State the formula that uses the value of E°cell & ΔG° to predict the feasibility of a reaction.

A

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

18
Q

Compare electrochemical & electrolysis cells. (4)

A

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.

19
Q

Electrolysis

A

The breaking down of a compound into its elements using an electric current.

20
Q

Factors affecting selective discharge in electrolysis.

A
  • 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
21
Q

Order of ease of oxidation at anode.

A

SO4 2- > NO3 - > Cl- > OH - > Br - > I -
- depends on concentration (refer to slides)

22
Q

Faraday’s Law.

A

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.

23
Q

Formula for Faraday’s Law.

A

F = Le
- F = Faraday constant (charge on 1 mole of e-)
- L = Avogadro constant
- e = charge on the electron

24
Q

What is the charge on 1 electron?

A

1.60 x 10^-19 C

25
Q

What is the value of Faraday’s constant?

A

96500 C

26
Q

Formula for the quantity of charge passed during electrolysis.

A

Q = It
- I = current in aWmperes, A
- t = time in seconds, s
- Q = quantity of charge in coulombs, C

27
Q

What is the value of Avogadro’s constant?

A

6.02 x 10^23