Aim # 23: What is the effect of concentration on the EMF of
a galvanic cell?
H.W. # 23
Study pp. 849 – 854 (up to “The Nernst Equation”)
Ans. Ques. p. 880 # 49 - 51, 59
I Determining when a redox reaction will occur spontaneously
For any redox reaction,
ε0 = ε0red(reduction) – ε0red(oxidation)
If ε0 > 0, the reaction occurs spontaneously
ε0 < 0, the reaction will not occur spontaneously
Problem: Will copper metal react spontaneously with acids?
Ans: The reaction in question is
Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
From the table of standard reduction potentials,
the half-reactions are
Cu2+ + 2e- → Cu ε0 = +.34 V
2H+ + 2e- → H2
ε0 = 0 V
ε0 = ε0(reduction) - ε0(oxidation)
ε0 = 0 V – (+.34 V)
ε0 = -.34 V
The reaction will not occur spontaneously.
WE NOW HAVE A BASIS FOR DEVELOPING AN
ACTIVITY SERIES!
II The relationship between EMF and free energy
EMF = potential difference (V) = energy (J)
charge (C)
energy = the ability to do work
ε = -w
WHY IS w NEGATIVE?
q
-w = qε
Note: ε = εmax and -w = wmax (theoretically)
We define: 1 faraday (F) = the charge on one mole
of electrons
1 faraday = 96,485 C/mol eTherefore, q = nF
where n = no. moles eF = 1 faraday of charge
wmax = ΔG
For a galvanic cell,
wmax = ΔG = -qεmax
or
ΔG = -nFε
and for standard conditions,
ΔG0 = -nFε0
See Ref. Tables
Problem: Use standard reduction potentials to calculate
ΔG0 for the reaction
3Sn2+(aq) + 2Al(s) → Sn(s) + 2Al3+(aq)
Is the reaction spontaneous?
Ans: The half-reactions are
ox: Al → Al3+ + 3ered: 3Sn2+ + 6e- → 3Sn
ε0 = -1.66 V
ε0 = -0.14 V
2Al → 2Al3+ + 6e3Sn2+ + 6e- → 3Sn
3Sn2+(aq) + 2Al(s) → 3Sn(s) + 2Al3+(aq)
n=6
ε0 = ε0red(cathode) - ε0red(anode)
ε0 = -0.14 v - (-1.66 V)
ε0 = +1.52 V
ΔG0 = -nFε0
ΔG0 = -(6mol e-)(96485 C/mol e-)(+1.52J/C)
ΔG0 = -879,943.2 J
ΔG0 = -880. kJ
The reaction is spontaneous
(ε0 is positive and ΔG0 is negative)
II The Effects of Concentration on ε
A. Under standard conditions (all concs. 1 M),
for the cell reaction
Cd(s) + 2Ag+(aq) → Cd2+(aq) + 2Ag(s)
ε0 = +0.80V – (-0.40V) = +1.20
How will increasing the conc. Of Ag+ affect ε0cell?
Increasing a reactant conc. favors the forward
reaction (Le Chatelier’s principle), ε0cell > 1.20 V
How will increasing the conc. Of Cd2+ affect
ε0cell?
Increasing a product conc. Opposes the forward
reaction, ε0cell < 1.20 V
B. Concentration Cell
Both half-cells contain the same components,
but different concentrations.
• The difference in conc. Is
the only factor that
produces a cell potential.
Therefore,
the voltage is small.
• In which direction will the
current flow?
Practice Problems
Zumdahl (8th ed.)
p. 864 # 61
p. 868 # 118, 119, 126