9. Redox Processes

Gain of oxygen loss of hydrogen loss of electrons

loss of oxygen gain of hydrogen gain of electrons

substance that causes oxidation to occur while itself is reduced

substance that causes reduction to occur while itself is oxidized

Anode - Cathode + oxidation happens at the anode

Anode + Cathode - oxidation at anode

• 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

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)

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

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

• 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

dG = -nFE n = number of moles F = Faraday constant, change per mole, 96500 C/mol

Metal (if metal is low in reactivity series) Hydrogen (if metal is high in the reactivity series)

Non-metal (if halide solution is concentrated) Oxygen / oxidation of the electrode

Cathode Cu2+(aq) + 2e- <—> Cu(s) +0.34 most positive H2O(l) + e- <—> 0.5H2(g) + OH-(aq) -0.83 Anode H2O(l) <—> 0.5O2(g) + 2H+(aq) + 2e- -1.23 Cu2+(aq) + H2O(l) –> Cu(s) + 0.5O2(g) + 2H+(aq) -0.89V

0.5O2(g) + 2H+(aq) + 2e- <—> H2O(l) +1.23 reverse –> anode H+(aq) + e- <—> 0.5H2(g) 0.00 cathode H2O(l) –> 0.5O2(g) + H2(g) E = -1.23V

Cathode H2O(l) + e- <—> 0.5H2(g) + OH-(aq) -0.83 most positive Na+(aq) + e- <—> Na(s) - 2.71 Anode H2O(l) <—> 0.5O2(g) + 2H+(aq) + 2e- - 1.23 most positive but for concentration of >25% Cl is preferred Cl-(aq) <—> 0.5Cl2(g) + e- -1.36 2H2O(l) + 2Cl(aq) –> Cl2(g) + H2(g) + 2OH-(aq) E=2.19V

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

• 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

Q = It Q = charge - the Faraday constant (96500 C/mol) I = current (measured in amps) t = time (measured in seconds)