Edexcel Chemistry - Topic 14: Redox II

  • Created by: Ryan C-S
  • Created on: 11-04-18 21:26

Oxidising and Reducing Agents

  • Oxidising Agents are Electron Acceptors e.g. Cl2 and end up being reduced
  • Reducing Agents are Electron Donors e.g. Mg and end up being oxidised
  • Fluorine is the best oxidising agent as it has the greatest tendancy to gain electrons due to its small size and high nuclear attraction
  • Caesium is the best reducing agent as it has the greatest tendancy to lose electrons due to its large size and high shielding which create a weaker nuclear attraction
  • Francium would be a stronger reducing agent but it unstable
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Electrochemical Cells

  • An electrochemical cell consists of two half cells connected by a salt bridge to generate a potential difference (voltage)
  • Half cells consist of a metal electrode and a solution of a compound containing the metal e.g. Cu and CuSO4
  • A potential difference forms when an external circuit is complete because one metal has a greater tendendcy to oxidise and release electrons than the other causing a build up of electrons on one electrode
    e.g. Zn has a greater tendency than Cu to oxidise so releases electrons establishing a p.d.
  • The salt bridge is used to connect up the circuit. The free moving ions conduct the charge
  • The salt bridge should be unreactive with the electrodes and solution
  • KNO3 is often used for this as opposed to KCl as the Chloride ions can form complexes with Cu2+ ions
  • A high resistance voltmeter is used to measure the voltage. A high resistance voltmeter must be used in order to stop the current flowing in the curcuit as this prevents the reactants from being used up
    Cu2+(aq) + 2e- --> Cu(s): Positive Electrode; electrons used up
    Zn(s) --> Zn2+(aq) + 2e- : Negative Electrode; electrons given off
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Electrochemical Cell Diagram

Image result for electrochemical cell diagram

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Cell Diagrams

  • Electrochemical cells can be represented by a cell diagram:
    Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
  • If a system doesn't include a metal that can act as an electrode, then a platinum electrode is used and included in the cell diagram
  • A platinum electrode is used because it is unreactive and can conduct electricity
    e.g. for the reaction Fe2+(aq) --> Fe3+(aq) + e- there is no solid conducting surface so a platinum electrode is used - || Fe3+(aq), Fe2+(aq) | Pt
  • A cell diagram the more oxidised species is closest to the || with the oxidised and reduced species seperated by a comma
  • If a half equation has multiple physical states then the solid vertical lines should be used between each state boundary
    e.g. Cl2(g) + 2e- --> 2Cl-(aq) ; || Cl2(g) | Cl-(aq) | Pt(s)
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Measuring Electrode Potentials

  • The absolute potential of a half cell must be calculated by measuring the potential difference between two electrodes
  • One of the half cells must have a known potential so the potential difference can between the two half cells can be measured
  • Each electrode is assigned a relative potential by linking it to a reference electrode (Standard Hydrogen Electrode) which is given a potential of zero volts
    H2(g) ⇌ 2H+(aq) + 2e-
    Pt(s) | H2(g) | H+(aq) || 
  • Platinum wire coated in porous platinum black (which acts as a catalyst) to allow the equilibrium to be achieve
    Standard Conditions are:
  • Temperature: 298K
  • Pressure: 100kPa
  • Solutions: 1.00mol dm-3
  • No current flowing
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Standard Hydrogen Electrode

H2(g) ⇌ 2H+(aq) + 2e-

Pt(s) | H2(g) | H+(aq) || Mg2+(aq) | Mg(s)

Image result for electrochemical cell diagram hydrogen

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Standard Electrode Potentials

  • Standard conditions are required because the position of the redox equilibrium will change with conditions
  • The Standard Hydrogen Electrode is difficult to use so a different easier electrode is used and calibrated against the Standard Hydrogen Electrode. These are called Secondary Standards
  • A common secondary standards include Silver|Silver Chloride
  • When an electrode system is connected to a standard hydrogen electrode, the potential difference measured is the standard electrode potential
  • Standard Electrode Potentials are found in data books with the more oxidised form on the left

Li+(aq) | Li(s) E = -3.03V

Li+(aq) + e- ⇌ Li(s) E = -3.03V

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Calculating Electromotive Force

  • Electromotive Force (emf) is a measure of the driving force of the whole cell i.e. the amount of work that is obtainable from the whole cell.
  • Internal resistance means that if a current flows, the potential difference or voltage obtained will be less than the emf
  • More Negative half equations will oxidise (<--)
    More Positive half equations will reduce (-->)
  • Ecell (emf) = Ered - Eox
  • The electrochemical series arranges redox equilibria in order of their standard electrode potentials
  • The more powerful reducing agents will be found at the most negative end of the electrochemical series and vice versa
  • The more negative electrodes will oxidise and go from left to right and allows a reactions feasibility to be determined
  • If the overall potential is positive, the reaction will be thermodynamically feasible
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Effect of Changing Conditions on Redox Equilibria

  • The effects of changing conditions on Ecell can be made by applying Le Chateliers Principle
  • Ecell is a measure of how far from the equilibrium the cell reaction lies. The more positive the Ecell, the more likely a reaction is to occur
  • If current is allowed to flow, the cell reaction will occur and the Ecell will fall to zero as the reaction will proceed until the reactant concentrations drop
  • If non standard conditions are used, a reaction may or may not go to completion

Concentration

  • Increasing concentrations of reactants would increase the Standard Electrode Potential due to Le Chateliers Principle

Temperature

  • Most cells are exothermic in the spontaneous direction so applying Le Chateliers principle to a temperature rise would result in a decrease in Standard Electrode Potential because the equilibrium would shift backwards in the endothermic direction
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Lead-Acid Cells

  • Rechargeable battery cells used to power the starter motor or lights in a car
  • PbSO4 + 2e- --> Pb + [SO4]2-  E = -0.036V
  • PbO2 + [SO4]2- + 4H+ + 2e- --> PbSO4 + 2H2O  E = +1.69V
  • PbO2 + Pb + 2[SO4]2- + 4H+ --> 2PbSO4 + 2H2O  E = +2.04V

Image result for lead acid battery diagram

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Lithium Battery

  • Non-rechargeable Battery cells that power laptops and mobile phone batteries
  • +ve Electrode: Li+ + MnO2 + e- --> LiMnO2
  • -ve Electrode: 2Li --> 2Li+ + 2e-
  • Li(s) | Li+(aq) || Li+(aq) | MnO2(s), LiMnO2(s) | Pt(s)
  • Lithium ions act as an electrolyte

Image result for lithium manganese cell diagram

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Hydrogen Fuel Cell

  • Hydrogen fuel cells are a renewable source of energy that maintains a constant voltage
  • +ve Electrode: 0.5 O2(g) + 2H+ + 2e- --> H2O(l)
  • -ve Electrode: H2(g) --> 2H+ + 2e-
  • Overall Reaction: H2(g) + 0.5 O2(g) --> H2O(l)
  • Hydrogen Fuel Cells are less polluting (releasing only water) and have a greater efficiency than standard electrochemical cells but are expensive, storing hydrogen is difficult and they have a limited lifetime so need regular replacement

Image result for hydrogen fuel cell diagram 

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Disproportionation Reactions

  • A disproportionation reaction is one where an element (species) is simulatenously oxidised and reduced
  • Copper(I) Iodide when reacting with sulphuric acid will disproportionate to Cu2+ and Cu metal
  • Cu+(aq) + e-  --> Cu(s)  E = +0.52V
  • Cu2+(aq) + e-  --> Cu+(aq) E = +0.15V
  • E = +0.52 - +0.15 = + 0.37V
  • If the overall potential has a positive value, the reaction will be thermodynamically feasible and disproportionation will occur
  • A reaction may be thermodynamically feasible but may have too high an activation energy or have a slow rate of reaction and may not go
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Redox Titrations

[MnO4]-(aq) + 8H+(aq) + 5Fe2+(aq) --> Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

End Point: Colourless --> Pink/Purple

Method

  • Add a known concentration of Fe2+ in FeSO4 to a conical flask using a pipette
  • Pipette excess H+ ions in sulphuric acid into the flask
  • Fill a burette with KMnO4
  • Open the tap over the flask and start swirling the flask
  • When the end point has been reached, turn off the tap
  • Record the change in levels on the burette to find out how much of the KMnO4 was used
  • Repeat until two concordant results have been achieved
  • Calculate the number of moles used and then the molarity (concentration) of the solution
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