CHEM 102 Class 13
Half-reaction
E0red (V)
Half-reaction
E0red (V)
Co3+ + e- → Co2+
+1.80
2 H + + 2 e- → H 2
0.00
Hg2+ + 2 e- → Hg
+0.85
Fe3+ + 3 e- → Fe
-0.04
Ag+ + e- → Ag
+0.80
Pb2+ + 2 e- → Pb
-0.13
I 2 + 2 e- → 2 I -
+0.54
Co2+ + 2 e- → Co
-0.29
Cu2+ + 2 e- → Cu
+0.15
Fe2+ + 2 e- → Fe
-0.41
Calculating E0cell
(1) Calculate E0cell for the reaction Ag+(aq) + Co2+(aq) → Ag(s) + Co3+(aq)
Half reactions:
Ag+ + e- → Ag reduction E0red = +0.80 V
Co2+ → Co3+ + 1 e- oxidation E0ox = -E0red = -(+1.80) = -1.80 V
E0cell = Eox + Ered = +0.80 V + (-1.80 V) = -1.00 V
(2) Calculate E0cell for the reaction Fe(s) + 2 H+(aq) → Fe2+(aq) + H2(g)
Half reactions:
Fe → Fe2+ + 2 e- oxidation E0ox = -E0red = -(-0.41 V) = +0.41 V
2 H+ + 2 e- → H2 reduction E0red = 0.00 V
E0cell = Eox + Ered = +0.41 V + 0.00 V = +0.41 V
(3) Calculate E0cell for the reaction 2 I-(aq) + 2 H+(aq) → I2(s) + H2(g)
Half reactions:
2 I- → I2 + 2 e- oxidation E0ox = -E0red = -(+0.54 V) = -0.54 V
2 H+ + 2 e- → H2 reduction E0red = 0.00 V
E0cell = Eox + Ered = -0.54 V + 0.00 V = -0.54 V
(4) Calculate E0cell for the reaction 2 Fe3+(aq) + 3 Pb(s) → 2 Fe(s) + 3 Pb2+(aq)
Half reactions:
2 Fe3+ + 6 e- → 2 Fe reduction E0red = -0.04 V
3 Pb → 3 Pb2+ + 6 e- oxidation E0ox = -E0red = -(-0.13 V) = +0.13 V
E0cell = Eox + Ered = +0.13 V + (-0.04 V) = +0.09 V
Cell diagrams/cell notation
(5) Write a cell diagram/cell notation for the reaction Cu2+(aq) + Fe(s) → Cu(s) +
Fe2+(aq)
Remember, the anode is written on the left and the anode is the site of oxidation
reactions. The oxidation reaction therefore appears first. The oxidation reaction is Fe(s)
→ Fe2+(aq) + 2 eFe(s)|Fe+2+(aq)||Cu2+(aq)|Cu(s)
(6) Write a cell diagram/cell notation for the reaction Co3+(aq) + Ag(s) → Co2+(aq) +
Ag+(aq)
The anode is written on the left and the anode is the site of oxidation reactions. The
oxidation reaction therefore appears first. The oxidation reaction is Ag(s) → Ag+(aq) + eAg(s)|Ag+(aq)||Co2+(aq), Co3+(aq)|Pt
Reaction Spontaneity
(7) Predict whether the reaction Co3+(aq) + Ag(s) → Co2+(aq) + Ag+(aq) will be
spontaneous as written by comparing E0red values
There are several ways to predict spontaneity. Two of the simplest are to calculate E0cell
(all reactions with a positive E0cell are spontaneous) or to compare the half-cell reduction
potentials (the half reaction with the most positive E0red goes spontaneously as a
reduction, the other is forced to run as an oxidation)
Looking at the reaction we can see that Co3+ + 1 e- → Co2+ has the most positive E0red
and this is how it’s written in the equation so the reaction is spontaneous. E0cell is +1.00
V
(8) Predict whether the reaction Pb2+(aq) + Cu(s) → Pb(s) + Cu2+(aq) will be spontaneous
as written by comparing E0red values
Looking at the reaction we can see that Cu2+ + 2 e- → Cu has the most positive E0red but
it’s written in the equation as the opposite (the half reaction is Cu → Cu2+ + 2 e-). E0cell is
-0.28 V
Calculating ΔG0rxn
(9) Calculate ΔG0rxn for the reaction Co2+(aq) + Fe(s) → Co(s) + Fe2+(aq) to decide
whether it is spontaneous.
The equation is ΔG0rxn = -n·F·E0cell where F = 96485 C/mol. First we need to calculate
E0cell. The half-reactions are:
Co2+(aq) + 2 e- → Co(s) reduction E0red = -0.29 V
Fe(s) → Fe2+(aq) + 2 e- oxidation E0ox = -E0red = -(-0.41 V) = +0.41 V
E0cell = E0ox + E0red = +0.41 V + (-0.29 V) = +0.12 V
ΔG0rxn = -2 mols e- x 96,485 C/mol x (+0.12 J/C) = -23,156 J = -23.2 kJ and the reaction
is spontaneous (we could also tell this from the positive E0cell)
(10) Calculate ΔG0rxn for the reaction 3 Co2+(aq) + 2 Fe(s) → 3 Co(s) + 2 Fe3+(aq) to
decide whether it is spontaneous.
The equation is ΔG0rxn = -n·F·E0cell where F = 96485 C/mol. First we need to calculate
E0cell. The half-reactions are:
3 Co2+(aq) + 6 e- → 3 Co(s) reduction E0red = -0.29 V (note, not 3 x -0.29 V!)
2 Fe(s) → 2 Fe3+(aq) + 6 e- oxidation E0ox = -E0red = -(-0.04 V) = +0.04 V
E0cell = E0ox + E0red = +0.04 V + (-0.29 V) = -0.25 V
ΔG0rxn = -6 mols e- x 96,485 C/mol x (-0.25 J/C) = +144,727.5 J = +144.7 kJ and the
reaction is non-spontaneous (we could also tell this from the negative E0cell)