1. PPT - Electrode potentials
14/04/2017
Electrode Potentials
know the IUPAC convention for writing half-equations for electrode reactions.
Know and be able to use the conventional representation of cells.
Know that standard electrode potential, E , refers to conditions of 298 K, 100 kPa and 1.00 mol dm−3 solution of ions.
A2.CHEM
5.3 EXERCISE 2 - electrochemical cells

2. Zn2+(aq) e–  Zn(s)

4. Zn  Zn2+ + 2 e- oxidation Cu2+ + 2 e-  Cu reduction
- electrode
anode
oxidation
+ electrode
cathode
reduction
electron flow
At this electrode the metal loses electrons and so is oxidised to metal ions.
These electrons make the electrode negative.
At this electrode the metal ions gain electrons and so is reduced to metal atoms.
As electrons are used up, this makes the electrode positive. Zn Cu
Zn  Zn e-
oxidation
Cu e-  Cu
reduction

9. Standard Conditions Concentration
1.0 mol dm-3 (ions involved in ½ equation)
Temperature
298 K
Pressure
100 kPa (if gases involved in ½ equation)
Current
Zero (use high resistance voltmeter)

10. S tandard
H ydrogen
E lectrode

11. Emf = E = Eright - Eleft

14. Pt(s) | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)

15. CELL DIAGRAMS Zn Zn2+ Cu2+ Cu
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = V put on the LHS
ZINC IS IN CONTACT THE SOLUTIONS A SOLUTION OF
WITH A SOLUTION ARE JOINED VIA A COPPER IONS IN
OF ZINC IONS SALT BRIDGE TO CONTACT WITH COPPER
Zn Zn Cu2+ Cu

16. + _ CELL DIAGRAMS Zn Zn2+ Cu2+ Cu
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = V put on the LHS
• Draw as shown… the cell reaction goes from left to right
• the zinc metal dissolves Zn(s) ——> Zn2+(aq) + 2e¯ OXIDATION
• copper is deposited Cu2+(aq) + 2e¯ ——> Cu(s) REDUCTION
• oxidation takes place at the anode
• reduction at the cathode _
Zn Zn Cu2+ Cu +

17. + _ CELL DIAGRAMS Zn Zn2+ Cu2+ Cu V
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = V put on the LHS
• Draw as shown… the electrons go round the external circuit from left to right
• electrons are released when zinc turns into zinc ions
• the electrons produced go round the external circuit to the copper
• electrons are picked up by copper ions and copper is deposited _
Zn Zn Cu2+ Cu + V

18. + _ CELL DIAGRAMS Zn Zn2+ Cu2+ Cu V
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = V put on the LHS
• Draw as shown… the cell voltage is E°(RHS) - E°(LHS) - it must be positive
cell voltage = V - (-0.76V) = V _
Zn Zn Cu2+ Cu + V

19. ROOR Ni(s) | Ni2+(aq) || Sn4+(aq), Sn2+(aq) | Pt(s)
K(s) | K+(aq) || Mg2+(aq) | Mg(s)

21. GOLDEN RULE The more +ve electrode gains electrons
(+ charge attracts electrons)

22. Electrodes with negative emf are better at releasing electrons (better reducing agents).

23. A2.CHEM
5.3 EXERCISE 2 - electrochemical cells

24. ELECTRODE POTENTIALS – Q1
Emf = Eright - Eleft
= Eright - 0
Eright = V

25. ELECTRODE POTENTIALS – Q2
Emf = Eright - Eleft
Emf =
Emf = V

26. ELECTRODE POTENTIALS – Q3
Emf = Eright - Eleft
Emf = (-0.76)
Emf = V

27. ELECTRODE POTENTIALS – Q4
Emf = Eright - Eleft
= Eleft
Eleft = = V

28. ELECTRODE POTENTIALS – Q5
Emf = Eright - Eleft
a) Emf = (-0.25) = V
b) Emf = = V
c) Emf = = V

29. ELECTRODE POTENTIALS – Q6
Emf = Eright - Eleft
a) Eright = = V
Ti3+(aq) + e-  Ti2+(aq)
b) Eleft = = V
K+(aq) + e-  K(aq)
c) Eright = = V
Ti3+(aq) + e-  Ti2+(aq)

30. ELECTRODE POTENTIALS – Q7
a) Cr(s) | Cr2+(aq) || Zn2+(aq) | Zn(s)
Emf = (-0.91) = V
b) Cu(s) |Cu2+(aq)|| Fe3+(aq),Fe2+(aq)| Pt(s)
Emf = = V
c) Pt(s) | Cl-(aq)| Cl2(g) || MnO4-(aq),H+(aq),Mn2+(aq)| Pt(s)
Emf = – 1.36 = V