Determination of Electrode Potentials Using Cell Potentials in Calculations
• Electrode potentials can be determined by • Cell potentials are additive
measurements versus the standard H-electrode or – If two reactions are added, their potentials are
other electrodes with known potentials added too
Example: Eocell = +0.46 V for the reaction: • Cell potentials are intensive properties – remain
independent of the system size
Cu(s) + 2Ag+ → Cu2+ + 2Ag(s)
– If a reaction (or a half-reaction) is multiplied by a
If Eo = +0.34 V for the Cu2+/Cu redox couple, what number, its potential remains the same
is Eo for the Ag+/Ag redox couple?
→ Split into half-reactions: Example:
Cu(s) → Cu2+ + 2e- EoCu = +0.34 V (anode, ox) Cu(s) + 2Ag+ → Cu2+ + 2Ag(s) (×3) Eocell = +0.46 V
Ag+ + e- → Ag(s) EoAg = ??? V (cathode, red) 3Ag(s) + Au3+ → 3Ag+ + Au(s) (×2) Eocell = +0.70 V
Eocell = EoAg – EoCu = EoAg – (+0.34) = +0.46 3Cu(s) + 2Au3+ → 3Cu2+ + 2Au(s)
⇒ EoAg = +0.46 + (+0.34) = +0.80 V Eocell = +0.46 + 0.70 = +1.16 V
Strengths of Oxidizing and Reducing Agents • Electrochemical series – an arrangement of the
• Eo values are always tabulated for reduction redox couples in order of decreasing reduction
potentials (Eo) → Appendix D
Ox + ne- → Red (Eo)
– The most positive Eos are at the top of the table
– Ox is an oxidizing agent; Red is a reducing agent
– The most negative Eos are at the bottom of the table
• Eo is a measure for the tendency of the half-reaction
to undergo reduction ⇒The strongest oxidizing agents (Ox) are at the top
of the table as reactants
⇒Higher (more positive) Eo means ⇒The strongest reducing agents (Red) are at the
– Greater tendency for reduction bottom of the table as products
– Lower tendency for oxidation • Every redox reaction is a sum of two half-reactions,
⇒Higher (more positive) Eo means one occurring as oxidation and another as reduction
– Stronger oxidizing agent (Ox) ← Ox is reduced Red1 → Ox1 + ne- Ox2 + ne- → Red2
– Weaker reducing agent (Red) ← Red is oxidized Red1 + Ox2 → Ox1 + Red2
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• In a spontaneous redox reaction, the stronger b) Can Cl2 oxidize H2O to O2 in acidic solution?
oxidizing and reducing agents react to produce the → Cl2/Cl- has higher Eo (Cl2/Cl- is above O2,H+/H2O)
weaker oxidizing and reducing agents ⇒ Cl2 is a stronger oxidizing agent than O2
Stronger Red1 + Stronger Ox2 → Weaker Ox1 + Weaker Red2 ⇒ Cl2 can oxidize H2O to O2 at standard conditions
Example: Given the following half-reactions: c) Write the spontaneous reaction between the Cl2/Cl-
Cl2(g) + 2e- → 2Cl- Eo = +1.36 V and Fe3+/Fe2+ redox couples and calculate its Eocell
Appendix D
O2(g) + 4H+ + 4e- → 2H2O(l) Eo = +1.23 V → Cl2/Cl- has the higher reduction potential (Eo)
Fe3+ + e- → Fe2+ Eo = +0.77 V
⇒ Cl2/Cl- undergoes reduction
Fe2+ + 2e- → Fe(s) Eo = –0.44 V ⇒ Fe3+/Fe2+ undergoes oxidation (reverse equation)
a) Rank the oxidizing and reducing agents by strength Cl2(g) + 2e- → 2Cl- (reduction) Eo = +1.36 V
+
→ Ox agents on the left; Red agents on the right Fe → Fe + e
2+ 3+ - ×2 (oxidation) Eo = +0.77 V
Oxidizing → (Top) Cl2 > O2 > Fe3+ > Fe2+ (Bottom) Cl2(g) + 2e- + 2Fe2+ → 2Cl- + 2Fe3+ + 2e-
Reducing → (Bottom) Fe > Fe2+ > H2O > Cl- (Top) Eocell = Eocath – Eoanod = +1.36 – (+0.77) = +0.59 V
d) Is the reaction of disproportionation (simultaneous Relative Reactivity of Metals
oxidation and reduction) of Fe2+ to Fe3+ and Fe(s) • The activity series of metals – ranks metals based
spontaneous at standard conditions? on their ability to displace H2 from acids or water or
→ Need the sign of Eocell displace each other’s ions in solution
⇒ Fe2+/Fe(s) undergoes reduction • Metals that can displace H2 from acids
⇒ Fe3+/Fe2+ undergoes oxidation (reverse equation) – The reduction of H+ from acids to H2 is given by the
standard hydrogen half-reaction
Fe2+ + 2e- → Fe(s) (reduction) Eo = –0.44 V
+ 2H+ + 2e- → H2(g) Eo = 0 V
Fe2+ → Fe3+ + e- ×2 (oxidation) Eo = +0.77 V – In order for this half-reaction to proceed as written,
3Fe2+ + 2e- → Fe(s) + 2Fe3+ + 2e- the metal must have lower reduction potential (the
Eocell = Eocath – Eoanod = –0.44 – (+0.77) = –1.21 V metal must be below H2/H+ in Appendix D)
⇒ If Eometal < 0, the metal can displace H2 from acids
⇒ Eocell < 0 → the reaction is non-spontaneous at
⇒ If Eometal > 0, the metal cannot displace H2
standard conditions
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Example: Can Fe and Cu be dissolved in HCl(aq)? • Metals that can displace H2 from water
→ Fe2+/Fe is below and Cu2+/Cu is above H2/H+ – The reduction of H2O to H2 is given by:
2H+ + 2e- → H2(g) (reduction) Eo = 0.00 V 2H2O(l) + 2e- → H2(g) + 2OH- E = -0.42 V
+
Fe(s) → Fe2+ + 2e- (oxidation) Eo = –0.44 V – The value of E is for pH = 7 (nonstandard state)
2H+ + 2e- + Fe(s) → H2(g) + Fe2+ + 2e- ⇒ Metals that are below H2O/H2,OH- in Appendix D
can displace H2 from water at standard conditions
Eocell = Eocath – Eoanod = 0.00 – (–0.44) = +0.44 V
⇒ Metals that have Eometal < -0.42 can displace H2
⇒ Eocell > 0 → spontaneous (Fe dissolves in HCl) from water at pH = 7
2H+ + 2e- → H2(g) (reduction) Eo = 0.00 V Example: Potassium, K, dissolves readily in water
+
Cu(s) → Cu2+ + 2e- (oxidation) Eo = +0.34 V 2H2O(l) + 2e- → H2(g) + 2OH- (reduction) E = –0.42 V
2H+ + 2e- + Cu(s) → H2(g) + Cu2+ + 2e- K(s) → K+ + e- ×2 (oxidation) Eo = –2.93 V
Eocell = Eocath – Eoanod = 0.00 – (+0.34) = –0.34 V 2H2O(l) + 2e- + 2K(s) → H2(g) + 2OH- + 2K+ + 2e-
⇒ Eocell < 0 → non-spontaneous (Cu doesn’t dissolve) Eocell = –0.42 – (–2.93) = +2.51 V > 0 (spontaneous)
