Redox

Electrochemical Cells The movement of electrons in a redox reaction can be used to generate a useful electric current. The principle is to separate the oxidation and reduction reactions and then the electrons are forced to travel around a circuit.

Electrochemical Cells Two electrodes are used to allow electrons to travel into or out of each half-reaction. These must be electrically conducting. If the half-reaction involves a metal, then this can be used for an electrode. Otherwise an inert metal is used (usually platinum). The redox reaction also requires movement of ions between the two half-reactions. This is achieved using a salt bridge.

The salt bridge contains a saturated solution of KCl or KNO 3 (usually soaked into agar jelly or absorbent paper)

The salt bridge allows ions to flow whilst keeping the two electrolyte solutions apart.

In one half-reaction electrons are released and the electrode becomes negative (as a result of OXIDATION). In the other electrons are gained and the electrode becomes positive (as a result of REDUCTION).

The electrode where oxidation takes place is called the anode. The electrode where reduction takes place is called the cathode. In an electrochemical cell the anode is negative (this can cause confusion!).

The displacement reaction between zinc and copper can be used in an electrochemical cell. Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) The half-reaction that is oxidation is: Zn (s) → Zn 2+ (aq) + 2e - This generates a negative potential. The half reaction that is reduction is: Cu 2+ (aq) + 2e - → Cu (s) This generates a positive potential.

A conventional cell diagram has the negative (oxidation) electrode on the left and the positive (reduction) electrode on the right.

For cell diagrams: ‖ represents a salt bridge. | represents a phase boundary. Different species in the same phase are separated by a comma. Oxidation on the left Reduction on the right

For cell diagrams: The conducting electrodes are at each end. The order of species is: reducedoxidised salt bridgeoxidisedreduced mnemonic: RO(S)OR

ROORROOR Zn (s) | Zn 2+ (aq) ‖ Cu 2+ (aq) | C u (s) Change:OXidationREDuction Sign:negativepositive -+-+

When a cell diagram is drawn conventionally, the direction of change for each electrode is left → right. Zn (s) | Zn 2+ (aq) ‖ Cu 2+ (aq) | C u (s) Change:OXidationREDuction Sign:negativepositive -+-+

Standard Electrode Potentials When an electrochemical cell is set up a potential difference (voltage) is established. A half-reaction which is more likely to lose electrons to the electrode (oxidise) produces a more negative potential. A half-reaction which is more likely to gain electrons from the electrode (reduce) produces a more positive potential.

Standard Electrode Potentials The factors that affect a cell potential are: Current drawn by the circuit. When zero current flows the potential is called the electromotive force (e.m.f.). Concentration of any solutions. Temperature. Pressure, if any gases are reacting. The standard e.m.f. is measured under standard conditions.

Standard Electrode Potentials Standard conditions are: Concentration of any solutions 1 mol dm -3 Temperature of 298K Pressure of 100kPa

Standard Hydrogen Electrode It is impossible to measure the potential of just one electrode, so a standard electrode has been chosen, against which all other half- cell potentials can be compared. This is the Standard Hydrogen Electrode.

Pt (s) | H 2(g) | H + (aq) This has, by definition, a standard electrode potential E o of 0·00V

For an electrochemical cell: E o cell = E o right – E o left To measure a standard electrode potential directly, a cell with the standard hydrogen electrode as the left hand electrode is used. It is time-consuming to set up a S.H.E., so usually a secondary standard with a very stable potential is used instead.

The electrochemical series This is a list of standard electrode potentials. The half-cell reactions are always given as reductions. The order of potentials varies from one version to another e.g.: most positive to most negative most negative to most positive alphabetical

The electrochemical series The most positive cell potentials are for species that are likely to be reduced (strongest oxidising agents). The most negative cell potentials are for species that are likely to be oxidised (strongest reducing agents).

The electrochemical series The cell reaction for the positive electrode goes in the forward direction. The cell reaction for the negative electrode goes in the reverse direction. (This would make more sense if equilibrium signs were used in the electrochemical series, but they are not!)

The electrochemical series This can be used to: calculate e.m.f. values for cells. predict the direction of redox reactions Often a shorthand version of a half equation is used when discussing elctrodes. e.g. Zn 2+ /Zn instead of Zn 2+ (aq) + 2e - → Zn (s) or Fe 3+ /Fe 2+ instead of Fe 3+ (aq) + e - → Fe 2+ (aq) These are called redox couples. Increasing oxidising power Increasing reducing power Reduction half-reactionE o / V F 2(g) + 2e - → 2F - (aq) MnO 4 2- (aq) + 4H + (aq) + 2e - → MnO 2(s) + 2H 2 O (l) MnO 4 - (aq) + 8H + (aq) + 5e - → Mn 2+ (aq) + 4H 2 O (l) Cl 2(g) + 2e - → 2Cl - (aq) Cr 2 O 7 2- (aq) + 14H + (aq) + 6e - → 2Cr 3+ (aq) + 7H 2 O (l) Br 2(l) + 2e - → 2Br - (aq) Ag + (aq) + e - → Ag (s) Fe 3+ (aq) + e - → Fe 2+ (aq) MnO 4 - (aq) + e - → MnO 4 2- (aq) I 2(s) + 2e - → 2I - (aq) Cu 2+ (aq) + 2e - → Cu (s) Hg 2 Cl 2(aq) + 2e - → 2Hg (l) + 2Cl - (aq) AgCl (s) + e - → Ag (s) + Cl - (aq) H + (aq) + 2e - → H 2(g) 0.00 Pb 2+ (aq) + 2e - → Pb (s) Sn 2+ (aq) + 2e - → Sn (s) V 3+ (aq) + e - → V 2+ (aq) Fe 2+ (aq) + 2e - → Fe (s) Zn 2+ (aq) + 2e - → Zn (s) Al 3+ (aq) + 3e - → Al (s) Mg 2+ (aq) + 2e - → Mg (s) Na + (aq) + e - → Na (s) Ca 2+ (aq) + 2e - → Ca (s) K + (aq) + e - → K (s) Li + (aq) + e - → Li (s) -3.05

Increasing oxidising power Increasing reducing power Reduction half-reactionE o / V 1 F 2(g) + 2e - → 2F - (aq) MnO 4 2- (aq) + 4H + (aq) + 2e - → MnO 2(s) + 2H 2 O (l) MnO 4 - (aq) + 8H + (aq) + 5e - → Mn 2+ (aq) + 4H 2 O (l) Cl 2(g) + 2e - → 2Cl - (aq) Cr 2 O 7 2- (aq) + 14H + (aq) + 6e - → 2Cr 3+ (aq) + 7H 2 O (l) Br 2(l) + 2e - → 2Br - (aq) Ag + (aq) + e - → Ag (s) Fe 3+ (aq) + e - → Fe 2+ (aq) MnO 4 - (aq) + e - → MnO 4 2- (aq) I 2(s) + 2e - → 2I - (aq) Cu 2+ (aq) + 2e - → Cu (s) Hg 2 Cl 2(aq) + 2e - → 2Hg (l) + 2Cl - (aq) AgCl (s) + e - → Ag (s) + Cl - (aq) H + (aq) + 2e - → H 2(g) Pb 2+ (aq) + 2e - → Pb (s) Sn 2+ (aq) + 2e - → Sn (s) V 3+ (aq) + e - → V 2+ (aq) Fe 2+ (aq) + 2e - → Fe (s) Zn 2+ (aq) + 2e - → Zn (s) Al 3+ (aq) + 3e - → Al (s) Mg 2+ (aq) + 2e - → Mg (s) Na + (aq) + e - → Na (s) Ca 2+ (aq) + 2e - → Ca (s) K + (aq) + e - → K (s) Li + (aq) + e - → Li (s) -3.05

Predicting the direction of a redox reaction The reaction at each electrode proceeds in the forward direction, only if the cell potential is positive. This means that the most positive electrode reaction is in the forwards direction and the least positive in the reverse direction.

Predicting the direction of a redox reaction e.g. Use the electrochemical series to predict whether Cu 2+ will oxidise Fe to Fe 2+. Answer: E o values Cu 2+ +2e - → Cu +0.34V E o values Fe e - → Fe-0.44V The electrode potential for the Cu 2+ /Cu couple is more positive than Fe 2+ /Fe, so, yes, copper(II) ions can oxidise Fe to Fe 2+. The diagram for the change in the question is: Fe (s) | Fe 2+ (aq) ‖ ‖Cu 2+ (aq) | Cu (s) E cell = E r – E l = (-0.44) = +0.78V

Predicting the direction of a redox reaction e.g. Use the electrochemical series to predict whether Cu 2+ will oxidise Fe 2+ to Fe 3+. Answer: E o value Cu 2+ +2e - → Cu +0.34V E o value Fe 3+ + e - → Fe V The electrode potential for the Fe 3+ /Fe 2+ couple is more positive than Cu 2+ /Cu, so, no, copper(II) ions cannot oxidise Fe 2+ to Fe 3+. The diagram for the change in the question is: Pt (s) | Fe (s) 2+, Fe 3+ (aq) ‖ Cu 2+ (aq) | Cu (s) E cell = E r – E l = (+0.77) = -0.43V

Predicting e.m.f. values and reactions that take place for electrochemical cells Step 1: draw the conventional cell diagram for the cell (remember: negative electrode on the left, ROOR) Step 2: Calculate the emf using E cell = E r – E l (it will be positive) Step 3: Change the direction for the oxidation (negative/left) electrode Step 4: Combine the two half equations ensuring that the electrons cancel out.

Predicting e.m.f. values and reactions that take place for electrochemical cells Example a)Predict whether magnesium will reduce vanadium(III) ions to vanadium(II) ions. b)Draw a conventional cell diagram for the standard cell c)Determine its e.m.f. d)Write an equation for the redox reaction.

Predicting e.m.f. values and reactions that take place for electrochemical cells Example a)Predict whether magnesium will reduce vanadium(II) ions. E o value Mg 2+ +2e - → Mg -2.36V E o value V 3+ + e - → V V The E o for V 3+ /V 2+ is more positive than Mg 2+ /Mg, magnesium is the more powerful reducing agent and it will reduce V 3+.

b)Draw a conventional cell diagram for the standard cell. (Remember: negative on left, ROOR) E o value Mg 2+ +2e - → Mg -2.36V E o value V 3+ + e - → V V Mg(s) | Mg (aq) 2+ ‖ V 3+ (aq), V 2+ (aq) | Pt (s)

E o value Mg 2+ +2e - → Mg -2.36V E o value V 3+ + e - → V V c)Determine its e.m.f. Mg(s) | Mg (aq) 2+ ‖ V 3+ (aq), V 2+ (aq) | Pt (s) E cell = E r – E l = – (-2.36) = V

E o value Mg 2+ +2e - → Mg -2.36V E o value V 3+ + e - → V V Mg (s) | Mg (aq) 2+ ‖ V 3+ (aq), V 2+ (aq) | Pt (s) d)Write an equation for the redox reaction. The cell e.m.f. is positive, so the direction of change is left to right for each electrode. The half equation for Mg 2+ /Mg is reversed. The V 3+ /V 2+ half equation is doubled.

E o value Mg 2+ +2e - → Mg -2.36V E o value V 3+ + e - → V V d)Write an equation for the redox reaction. The cell e.m.f. is positive, so the direction of change is left to right for each electrode. The half equation for Mg 2+ /Mg is reversed. The V 3+ /V 2+ half equation is doubled. Mg (s) → Mg 2+ (aq) +2e - 2V 3+ (aq) + 2e - → 2V 2+ (aq) 2V 3+ (aq) + Mg (s) → 2V 2+ (aq) + Mg 2+ (aq)

Example a)Predict whether manganate(VI) ions will disproportionate manganate(VII) and manganese(IV) oxide. b)Draw a conventional cell diagram for the standard cell c)Determine its e.m.f. d)Write an equation for the redox reaction.

Example a)Predict whether manganate(VI) ions will disproportionate manganate(VII) and manganese(IV) oxide. E o : MnO e - → MnO V E o : MnO H + + 2e - → MnO 2 + 2H 2 O+1.55V The E o for is MnO 4 2- / MnO 2 more positive than MnO 4 - / MnO 4 2-, indicating that MnO 4 2- will be both oxidised and reduced.

Example b)Draw a conventional cell diagram for the standard cell. (Remember: negative on left, ROOR) E o : MnO e - → MnO V E o : MnO H + + 2e - → MnO 2 + 2H 2 O+1.55V Pt (s) | MnO 4 2- (aq), MnO 4 - (aq) ‖ MnO 4 2- (aq),H + (aq) | MnO 2(s), Pt (s)

Example c)Determine its e.m.f. E o : MnO e - → MnO V E o : MnO H + + 2e - → MnO 2 + 2H 2 O+1.55V Pt (s) | MnO 4 2- (aq), MnO 4 - (aq) ‖ MnO 4 2- (aq), H + (aq) | MnO 2(s), Pt (s) E cell = E r – E l = – (+0.60) = V

E o : MnO e - → MnO V E o : MnO H + + 2e - → MnO 2 + 2H 2 O+1.55V Pt (s) | MnO 4 2- (aq), MnO 4 - (aq) ‖ MnO 4 2- (aq), H + (aq) | MnO 2(s), Pt (s) d)Write an equation for the redox reaction. The cell e.m.f. is positive, so the direction of change is left to right for each electrode. The half equation for MnO 4 - /MnO 4 2- is reversed. The MnO 4 - /MnO 4 2- half equation is doubled. 2MnO 4 2- (aq) → 2MnO 4 - (aq) +2e - MnO 4 2- (aq) +4H + (aq) + 2e - → MnO 2(s) + 2H 2 O (l)

The cell e.m.f. is positive, so the direction of change is left to right for each electrode. The half equation for MnO 4 - /MnO 4 2- is reversed. The MnO 4 - /MnO 4 2- half equation is doubled. 2MnO 4 2- (aq) → 2MnO 4 - (aq) +2e - MnO 4 2- (aq) +4H + (aq) + 2e - → MnO 2(s) + 2H 2 O (l) 3MnO 4 2- (aq) + 4H + (aq) → 2MnO 4 - (aq) + MnO 2(s) + 2H 2 O (l)

Effects of changing conditions on cell e.m.f. When conditions change away from standard conditions, the half-reaction at each electrode should be treated as an equilibrium process and then LeChatelier’s principle can be applied.

Effects of changing conditions on cell e.m.f. When a temperature is changed, the equilibrium position shifts to oppose this. If more electrons are released the electrode potential becomes more negative, and if more electrons are gained (reduction) the electrode potential becomes more positive. Most of the electrode half-reactions are exothermic (not entropy driven), so high temperatures result in decreased e.m.f.s.

Effects of changing conditions on cell e.m.f. Example: calculate the e.m.f. of the zinc/copper cell. Zn (s) | Zn 2+ (aq) ‖ Cu 2+ (aq) | C u (s) Zn 2+ (aq) + 2e - → Zn (s) E o = -0.76V Cu 2+ (aq) + 2e - → Cu (s) E o = +0.34V E cell = E r – E l = – (-0.76) = V

Effects of changing conditions on cell e.m.f. How would the cell potential change if the concentration of Zn 2+ ions was increased? Consider the reaction at the zinc electrode: Zn (s) ⇌ Zn 2+ (aq) + 2e - (oxidation) The higher concentration of Zn 2+ is opposed by the equilibrium shifting to the left, so more electrons are gained and the electrode potential becomes more positive.

Zn (s) ⇌ Zn 2+ (aq) + 2e - (oxidation) The higher concentration of Zn 2+ is opposed by the equilibrium shifting to the left, so more electrons are gained and the potential becomes less negative (more positive). E cell = E r – E l originally E cell = – (-0.76) = V the magnitude of the zinc electrode potential reduces, resulting in a less positive (smaller) e.m.f.

Commercial Electrochemical Cells Primary cells: cannot be recharged Secondary cells: use reversible redox reactions, so they can be recharged using a current in the opposite direction to normal discharge. Fuel cells: use a constant supply of a fuel and oxygen (usually the air) to generate electricity from a redox reaction. This means that the potential difference (voltage) is constant.

Primary Cells Dry cell. Zn 2+ (aq) + 2 e − → Zn(s) E° = −0.76 V MnO 2 (s) + H 2 O(l) + e − → MnO(OH)(s) + OH − (aq) E° = V

Primary Cells Alkaline cell. Zn(OH) 2(s) + 2e − → Zn (s) + 2OH − (aq) E° = V MnO 2(s) + H 2 O (l) + e − → MnO(OH) (s) + OH − (aq) E°= V

Primary Cells Mercury cell. HgO (s) + H 2 O (l) + 2e − → Hg (l) + 2OH − (aq) E° = V Zn(OH) 2(s) + 2e − → Zn (s) + 2OH − (aq) E° = V

Primary Cells Lithium cell. Li + + MnO 2(s) + e − → LiMnO 2(s) E° = V Li + + e − → Li (s) E° = V

Secondary Cells Lead Acid cell PbO 2(s) + 3H + (aq) + HSO 4 - (aq) + 2e − → PbSO 4(s) + 2H 2 O (l) E° = V PbSO 4(s) + H + (aq) + 2e − → Pb (s) + HSO 4 - (aq) E° = V

Secondary Cells Nickel Cadmium cell NiO(OH) (s) + H 2 O (l) + e − → Ni(OH) 2(s) + OH - (aq) E° = V Cd(OH) 2(s) + 2e − → Cd (s) + 2OH - (aq) E° = V

Secondary Cells Lithium Ion cell Li + + CoO 2 + e − → LiCoO 2 E° = V Li + + e − → Li (s) E° = V

Secondary Cells Charge and discharge – When a secondary cell is discharging, the redox reaction proceeds as usual. When a charging current is supplied the polarity of the electrodes is forced to change and the reaction reverses.

Fuel Cells Fuel cells use a supply of a fuel and oxygen. As long as both are supplied the cell will operate at a constant voltage. Fuel cells convert chemical potential enery to electrical energy with far greater efficiency than indirect methods where a lot of energy is lost as heat.

Fuel Cells The hydrogen-oxygen fuel cell is probably the simplest. O 2(g) + 2H 2 O (l) + 4e- → 4OH - (aq) E o = +0.40V 2H 2 O (l) + 2e - → H 2(g) + 2OH - (aq) E o = -0.83V

Fuel Cells The hydrogen-oxygen fuel cell needs to be run at higher temperatures to provide a useful amount of current. This will reduce the cell e.m.f. (LeChatelier), so the pressure is adjusted to help compensate bar is a typical operating pressure.

Fuel Cells The hydrogen-oxygen fuel cell produces water as its exhaust gas. This could help to reduce CO 2 emissions. However care must be taken to include the emissions created in obtaining the hydrogen fuel.