General Chemistry II

Electrochemistry

Use this for “Find the voltage of Fe | Fe²⁺ (0.10 M) ‖ Cu²⁺ (0.0050 M) | Cu at 25 °C, where E° = 0.787 V.”

Diagram

Fe anode (−)[Fe²⁺] 0.1 MCu cathode (+)[Cu²⁺] 0.005 MKNO₃e⁻0.749 V0.720.740.760.780.80.82voltsE°EQ = 20: −(RT ÷ nF) ln Q = −0.0385 V
E° = 0.787 V at 1 M; here Q = [Fe²⁺] ÷ [Cu²⁺] = 20.
E
=
E°
−
(
RT
nF
)
ln
Q
E
=
0.787
V
−
(8.314
×
298.15
2 × 96,485
)
×
ln
20
E = 0.749 V
Dots: each beaker's ions in proportion (36 for the richer one).
This is an example. Type your own values to start a new problem.
E (V)
=
E° (V)
−
R
T (K)
F
ln
Q
canode
Anode ion concentration
example · M
ccathode
Cathode ion concentration
example · M

Formulas

Q
=
canode
ccathode
20
=
0.1
0.005
E
=
E
°
−
(8.314
×
T
n
× 96,485
)
×
ln(
Q
)
+0.749
=
0.787
−
(8.314
×
298.15
2 × 96,485
)
×
ln(20)

Assumptions

•
The cell is M | M²⁺ ‖ N²⁺ | N, both ions with the same charge (as Zn | Zn²⁺ ‖ Cu²⁺ | Cu), so Q = [anode ion] ÷ [cathode ion]. The solid metals are left out of Q.
•
E = E° − (RT ÷ nF) ln Q, with R = 8.314 J/(mol·K) and F = 96,485 C/mol. At 25 °C this is E° − (0.05916 ÷ n) log Q.
•
E° = E°(cathode) − E°(anode), from a table of standard reduction potentials; n is the electrons each half-reaction moves.
•
Temperatures are in kelvins (°C + 273.15): 25 °C is 298.15 K.

Step-by-step

WE KNOW
E° = +0.787 V, n = 2, T = 298.15 K, canode = 0.1 M, ccathode = 0.005 M
FIND
reaction quotient (Q), cell potential (E)
Step 1 · Find reaction quotient (Q)
Use Q = canode ÷ ccathode
The
metals
are
solids
and
stay
out
of
Q
:
divide
the
anode’s
ion
concentration
by
the
cathode’s.
Q
=
0.1
0.005
Q
=
20
Step 2 · Find cell potential (E)
Use E = E° − (8.314 × T ÷ (n × 96,485)) × ln(Q)
Find
RT
nF
in
volts,
multiply
it
by
ln
Q
,
and
take
that
from
E
°.
E
=
0.787
−
(8.314
×
298.15
2 × 96,485
)
×
ln(20)
E
=
0.787
−
0.01285
×
2.9957
E
=
0.787
−
0.03849
E
=
+0.749
V
→
E
>
0:
the
cell
runs
as
written
and
can
do
work.
Check
Put the numbers back in. Both sides should match.
20
=
0.1
0.005
✓
+0.7485
=
0.787
−
(8.314
×
298.15
2 × 96,485
)
×
ln(20)
✓
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