9. Redox Processes

Question 1

Oxidation

Answer 1

Gain of oxygen

loss of hydrogen

loss of electrons

Question 2

Reduction

Answer 2

loss of oxygen

gain of hydrogen

gain of electrons

Question 3

Oxidizing Agent

Answer 3

substance that causes oxidation to occur while itself is reduced

Question 4

Reducing Agent

Answer 4

substance that causes reduction to occur while itself is oxidized

Question 5

Voltaic Cell

Answer 5

Anode -

Cathode +

oxidation happens at the anode

Question 6

Electrolytic Cell

Answer 6

Anode +

Cathode -

oxidation at anode

Question 7

Inert Electrolysis

Answer 7

• If the metal is high in the reactivity series you will get hydrogen
• If the metal is low in the reactivity series you will get the metal
• if the halide solution is concentrated you will get the halogen (Cl, Br, I)
  
  • with other negative ions you will get oxygen

Question 8

Electromotive Force (EMF)

Answer 8

A voltaic cell produces a potential difference known as the electromotive force (EMF).

The cell potential or Electrode Potential (E) is measured by comparing it to a standard which is the standard hydrogen electrode (SHE)

Question 9

Standard Hydrogen Electrode

Answer 9

Potential difference created by 1 mole of hydrogen ions at 100kPa H₂ at 298K, which has a E value of 0.00V

• platinum electrode
• glass tube with holes to allow gas to escape

Question 10

The Electrochemical Series

Answer 10

• The more positive / larger E/v
  
  • reaction more likely to work –> more powerful
  • will reverse an equation with a lower value

Question 11

Predicting with E value

Answer 11

• E value can be used to predict the feasibility of redox and cell reactions
• In theory any redox reaction with a positive E value will work
• In practice, it proceeds if the E Value of the reaction is greater than +0.40V
  
  • ​an equation with a more positive E value reverse a less positive one

Question 12

Gibbs and Faraday constant

Answer 12

dG = -nFE

_{n = number of moles}

_{F = Faraday constant, change per mole, 96500 C/mol}

Question 13

Products formed at the Cathode

Answer 13

Metal (if metal is low in reactivity series)

Hydrogen (if metal is high in the reactivity series)

Question 14

Products formed at the Anode

Answer 14

Non-metal (if halide solution is concentrated)

Oxygen / oxidation of the electrode

Question 15

Electrolysis of CuSO_4(aq)

Answer 15

Cathode

Cu²⁺_(aq) + 2e^- <—> Cu_(s) +0.34 most positive

H₂O_(l) + e^- <—> 0.5H_2(g) + OH^-_(aq) -0.83

Anode

H₂O_(l) <—> 0.5O_2(g) + 2H⁺_(aq) + 2e^- -1.23

Cu²⁺_(aq) + H₂O_(l) –> Cu_(s) + 0.5O_2(g) + 2H⁺_(aq) -0.89V

Question 16

Electrolysis of Water

Answer 16

0.5O_2(g) + 2H⁺_(aq) + 2e^- <—> H₂O_(l) +1.23 _{reverse –> anode}

H⁺_(aq) + e^- <—> 0.5H_2(g) 0.00 _cathode

H₂O_(l) –> 0.5O_2(g) + H_2(g) E = -1.23V

Question 17

Electrolysis of NaCl_(aq)

Answer 17

Cathode

H₂O_(l) + e^- <—> 0.5H_2(g) + OH^-_(aq) -0.83 _{most positive}

Na⁺_(aq) + e^- <—> Na_(s) - 2.71

Anode

H₂O_(l) <—> 0.5O_2(g) + 2H⁺_(aq) + 2e^- - 1.23

_{most positive but for concentration of >25% Cl is preferred}

Cl^-_(aq) <—> 0.5Cl_2(g) + e^- -1.36

2H₂O_(l) + 2Cl_(aq) –> Cl_2(g) + H_2(g) + 2OH^-_(aq) E=2.19V

Question 18

Factors affecting the amount of products in electrolysis

Answer 18

• Increased charge on ion decreases product
• Increased current flowing through the circuit increases product
• Increased duration of electrolysis increases product

Question 19

Electroplating

Answer 19

• Item to be coated is placed on the cathode
• The anode may be a metal block, often the solution is cyanide

Applications

• Jewellery - cheaper
• Tin cans - prevents rusting
• Galvanizing - prevents rusting

Question 20

Quantitative Electrolysis

Answer 20

Q = It

_{Q = charge - the Faraday constant (96500 C/mol)}

_{I = current (measured in amps)}

_{t = time (measured in seconds)}