SME
General Chemistry II · Problem type
Voltage of a concentration cell
Use this for “One silver electrode sits in 0.0050 M Ag⁺ and another in 0.80 M Ag⁺, joined by a salt bridge. Find the cell’s voltage at 25 °C.”
Diagram
One metal both sides: a concentration cell, E° = 0.
The dilute side is oxidized, so its ions grow.
E
=
E°
−
(
RT
nF
)
ln
Q
E
=
0
V
−
(8.314
×
298.15
1 × 96,485
)
×
ln
0.00625
E = 0.13 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)
=
0.05916
V
n
log₁₀
cconc (M)
cdil (M)
Formulas
E
=
(
0.05916
n
)
×
log₁₀(
cconc
cdil
)
0.13
=
(
0.05916
1
)
×
log₁₀(
0.8
0.005
)
Assumptions
•
Both electrodes are the same metal in its own ions, so E° = 0. Only the difference in concentration drives the cell.
•
The dilute side is the anode: its metal dissolves and its ions grow. On the concentrated side the ions plate out, until both sides match and E = 0.
•
At 25 °C, RT ln 10 ÷ F = 0.05916 V, so E = (0.05916 ÷ n) log₁₀(cconc ÷ cdil), with n the charge on the ion.
Step-by-step
WE KNOW
n = 1, cdil = 0.005 M, cconc = 0.8 M
FIND
cell potential (E)
Step 1 · Find cell potential (E)
Use E = (0.05916 ÷ n) × log₁₀(cconc ÷ cdil)
E
°
=
0,
so
the
Nernst
equation
leaves
only
the
log
term.
Q
=
cdil
cconc
,
and
−log₁₀
Q
=
log₁₀(
cconc
cdil
).
E
=
(
0.05916
1
)
×
log₁₀(
0.8
0.005
)
E
=
0.05916
×
log₁₀(160)
E
=
0.05916
×
2.204
E
=
0.13
V
Check
Put the numbers back in. Both sides should match.
0.13
=
(
0.05916
1
)
×
log₁₀(
0.8
0.005
)
✓
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Related lessons
Electrochemistry
Main lesson
Free energy and K from a cell potential
Use this for “A cell has E° = 0.462 V and moves 2 electrons. Find ΔG° and K at 25 °C.”
Electrolysis: mass plated by a current
Use this for “A current of 2.50 A runs for 45 min through copper(II) sulfate. What mass of copper plates onto the cathode?”, or the time or current a mass needs.