Module 5.2.3- Redox and electrode potentials

• Oxidising agent


  A reagent that oxidises (takes electrons from) another species. It becomes reduced itself
• Reducing agent


  A reagent that reduces (adds electrons to) another species. It becomes oxidised itself
• Construction of redox equations using half-equations


  Balance electrons on both electrons

  Add and cancel electrons

  
• Construction of redox equations from oxidation numbers

  
• Writing a half-equation


  Assign oxidation number and the change

  Balance the electrons

  Balance any remaining atoms (oxygen) by adding H2O. Then add H+ to balance the equation.

  
• Manganate (VII) titrations

  MnO4- reduced

  1) standard solution added to Burnett

  2) add volume of solution being analysed and an excess of dilute sulfuric acid
• 2 examples of Manganate (VII) titrations

  iron (II) ions, Fe2+ ions

  ethanedioic acid (COOH)2
• Working out full equation between Manganate (VII) ions and Iron (II) ions


  R: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

  O: Fe2+ → Fe3+ + e-

  Balanced equation: MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ 4H2O
• Working out percentage purity of an iron(II) compound steps (of known moles of water e.g. Fe2O3:7H2O)

  Work out moles of known solution

  Work out the molar ratio if Fe2+ and determine the amount of Fe2+ that has reacted

  Scale up moles (if needed)

  Find mass of Fe2O3:7H2O by multiplying moles by Mr

  Work out percentage of original
• Determine the value of x and formula of (COOH)2:xH2O (of unknown moles of water e.g. (COOH)2:xH2O)

  Work out moles of MnO4- that has reacted

  Work out molar ratio between MnO4- and (COOH)2 and determine the moles of (COOH)2

  Now work out the Mr of (COOH)2:xH2O by doing mass/moles (that you just calculated)

  The Mr of (COOH)2 is 90 so take this away from the Mr you just calculated and then divide by 18 (16 + 2 in H2O) to find out x.
• Iodine/thiosulfate redox balanced equation

  
• Analysis of oxidising agents using Na2S2O3


  Na2S2O3 in burette with the oxidising agent in the conical flask

  Add excess potassium iodide to the flask which will turn it a brown - yellow colour (due to I- -> I2)

  Titrate this solution with Na2S2O3 and brown colour will slowly fade but to observe this more clearly add starch and you will observe a blue - black colour

  As end-point is being reached a pale-straw colour must be observed when titrating with Na2S2O3
• Why is starch added towards the end-point of a reaction


  A deep blue - black colour is formed to help assist with the identification of an end-point

  Once the blue-black colour is faded as its being titrated with sodium thiosulfate and you are able to observe a pale straw colour then you have reached the end point
• Simple explanation of a sodium thiosulphate titration with potassium iodide


  Sodium Thiosulfate is in the burette

  Oxidising agent in conical flask

  Excess potassium iodide is added to the flask

  It is now yellow brown due to iodine

  Titrate with Sodium Thiosulfate

  Add starch

  It is now blue - black

  Continue adding Sodium Thiosulfate until a pale straw colour is observed

  
• Analysis of brass

  0.500g sample of brass is reacted with concentrated HNO3 to form a solution containing Cu2+ and Zn2+ ions. The solution is then neutralised

  Excess KI(aq) is added and Cu2+ ions react with I- to form I2

  Iodine is then titrated with Na2S2O3

  
• Standard electrode potential


  The e.m.f. of a half cell compared with a standard hydrogen electrode, measured at 298 K with solution concentrations of 1 mol dm^-3 and gas pressure of 1 atm(100kPa)

  
• The larger (more positive) the Eꝋ value the more...


  ...feasible a reaction is.

  it will be reduced from left to right
• Half-cell


  involves an element in 2 oxidation states

  a simple example is a strip of metal in a solution of its own ions e.g. Cu(s) and Cu2+
• What electrode do you use when an ion half cell contains ions of the same element in a different oxidation state?

  e.g. Fe3+ + e- <=> Fe2+

  Platinum electrode (or an inert metal electrode)
• Types of half cells which use a

  platinum electrode

  • gases in contact with solutions of their ionse.g. Cl2(g) + 2e- <=> 2Cl-
  • solutions of ions in two different oxidation states e.g. Fe3+ + e- <=> Fe2+
  • solutions of oxidising agents in acid solution e.g. MnO4- + 8H+ + 5e- <=> Mn2+ + 4H2O
• Comparing two half equations, the one with the more positive Eꝋ value...


  ...gets reduced and the other half equation will get oxidised
• The negative electrode is the electrode with the _______ reactive metal which ______ electrons and is ________

  
  

  The positive electrode is the electrode with the ______ reactive metal which _______ electrons and is __________

  The negative electrode is the electrode with the more reactive metal which loses electrons and is oxidised

  The positive electrode is the electrode with the less reactive metal which gains electrons and is reduced
• Standard hydrogen electrode

  Standard hydrogen electrode potential of 0.00V

  298 K

  Hydrogen gas at 1 atm[H+] =1 mol dm^-3

  Platinum electrode

  
• Combination of half cells setup- list of what to include

  -salt bridge (flow of ions)

  -positive and negative electrodes(flow of electrons)

  -voltmeter

  -[ions in solution] =1 mol dm^-3

  -298K

  -1 atm(100 kPa)

  
• How do you prepare a salt bridge


  Soak a strip of filter paper in a saturated solution of potassium nitrate, KNO3.
• The more negative the Eꝋ value is, the ________ the reactivity of a metal in losing electrons (oxidation)

  
  

  The more positive the Eꝋ value is, the ________ the reactivity of a non-metal in gaining electrons (reduction)

  The more negative the Eꝋ value is, the greater the reactivity of a metal in losing electrons

  The more positive the Eꝋ value is, the greater the reactivity of a non-metal in gaining electrons
• What does the voltage reading in a half-cell setup mean

  The difference between the 2 electrode potential values
• What must you be careful of when you are moving away from standard conditions of electrode potential

  E must be used and not Eꝋ
• What effect will concentration changes have on a half-cell's electrode potential value- EXAM TECHNIQUE


  1) [X] increases/decreases

  2) equilibrium shifts to the right / left

  3) therefore reducing /oxidising [X]

  4) and absorbing / releasing more electrons

  5) and making E more positive /negative

  6) if talking about cell potential, talk about the difference whether its smaller or larger
• Calculation of a standard cell potential by combining two standard electrode potentials

  Eꝋ (cell) =Eꝋ (positive electrode)-Eꝋ (negative electrode)
• Limitations of predictions using Eꝋ values:


  reaction rate many reactions have a very large activation energy which results in a slow rate of reaction

  give no indication of the rate of reaction
• Limitations of predictions using Eꝋ values:


  concentration if the concentration of a solution is not 1 mol dm^-3, then the value of the electrode potential will be different from the standard value

  (please consult the exam technique card on this)
• Primary cells


  non-rechargeable used once only electrical energy produced by oxidation and reduction at the electrodes reactions cannot be reversed eventually chemicals used up, voltage falls used for low current long storage devices e.g. smoke detectors
• Secondary cells


  rechargeable cell reaction reversed during recharging chemicals regenerated - can be used again e.g. car batteries lead-acid
• Fuel cells


  -these produce energy through a cell reaction that relies on external supplies of a fuel and an oxidant (oxygen)-these need to be supplied continuously - if the fuel or oxygen runs out, the cell fails-not need to be recharged
• What makes a fuel cell different from an ordinary cell? (COMMON QUESTION)

  A fuel cell uses energy from the reaction of a fuel with oxygen to generate a voltage
• Advantages of hydrogen fuel cells


  • can be produced from water
  • environmentally friendly
  • easier to store than electricity
  • non toxic
  • no CO2
  • high efficiency (45-65%)
• Alkali hydrogen fuel cell


  redox systems

  2 H2O + 2 e- <-> H2(g) + 2 OH- (Eꝋ= -0.83V)

  1/2 O2(g) + H2O(l) + 2 e- <-> 2 OH- (Eꝋ= 0.40V)

  overall:

  - same as acid with same Eꝋ cell (1.23V)

  H2(g) + 1/2 O2(g) --> H2O(l)(Eꝋcell= 1.23V)
• Acid hydrogen fuel cell


  redox systems

  2 H+ +2 e- <-> H2(g) (Eꝋ= 0.00V)

  1/2 O2(g) + 2 H+(aq) +2 e- <-> H2O(l) (Eꝋ = 1.23V)

  overall:

  - same as alkali with same Eꝋ cell (1.23V)

  H2(g) + 1/2 O2 --> H2O(l) (Eꝋ cell= 1.23V)