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For the cell Fe(s) mid Fe²⁺(aq) ∥ Cu²⁺(aq) mid Cu(s)… | StemPy Lab
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For the cell Fe(s) mid Fe²⁺(aq) ∥ Cu²⁺(aq) mid Cu(s) with E° = 0.78 V, which Nernst equation is correct at 298 K?
Chemistry
Medium
High School (11–12)
Electrochemistry › Galvanic Cells, Electrode Potential and the Nernst Equation
Board Exam
AP
A-Level
IB
Created by
StemPy Lab
For the cell
Fe
(
s
)
∣
Fe
2
+
(
a
q
)
∥
Cu
2
+
(
a
q
)
∣
Cu
(
s
)
\text{Fe}(s) \mid \text{Fe}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu}(s)
Fe
(
s
)
∣
Fe
2
+
(
a
q
)
∥
Cu
2
+
(
a
q
)
∣
Cu
(
s
)
with
E
∘
=
0.78
V
E^\circ = 0.78\ \text{V}
E
∘
=
0.78
V
, which Nernst equation is correct at
298
K
298\ \text{K}
298
K
?
E
=
0.78
V
−
0.0591
V
log
[
Fe
2
+
]
[
Cu
2
+
]
E = 0.78\ \text{V} - 0.0591\ \text{V}\log\dfrac{[\text{Fe}^{2+}]}{[\text{Cu}^{2+}]}
E
=
0.78
V
−
0.0591
V
lo
g
[
Cu
2
+
]
[
Fe
2
+
]
E
=
0.78
V
−
0.0591
V
2
log
[
Fe
2
+
]
[
Fe
]
[
Cu
2
+
]
[
Cu
]
E = 0.78\ \text{V} - \dfrac{0.0591\ \text{V}}{2}\log\dfrac{[\text{Fe}^{2+}][\text{Fe}]}{[\text{Cu}^{2+}][\text{Cu}]}
E
=
0.78
V
−
2
0.0591
V
lo
g
[
Cu
2
+
]
[
Cu
]
[
Fe
2
+
]
[
Fe
]
E
=
0.78
V
−
0.0591
V
2
log
[
Fe
2
+
]
[
Cu
2
+
]
E = 0.78\ \text{V} - \dfrac{0.0591\ \text{V}}{2}\log\dfrac{[\text{Fe}^{2+}]}{[\text{Cu}^{2+}]}
E
=
0.78
V
−
2
0.0591
V
lo
g
[
Cu
2
+
]
[
Fe
2
+
]
E
=
0.78
V
−
0.0591
V
2
log
[
Cu
2
+
]
[
Fe
2
+
]
E = 0.78\ \text{V} - \dfrac{0.0591\ \text{V}}{2}\log\dfrac{[\text{Cu}^{2+}]}{[\text{Fe}^{2+}]}
E
=
0.78
V
−
2
0.0591
V
lo
g
[
Fe
2
+
]
[
Cu
2
+
]
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← A reaction has a large positive E° and a very large K, yet a mixture of the reactants shows no change for years. What is the most likely reason?
The cell Fe(s) mid Fe²⁺(aq) ∥ Cu²⁺(aq) mid Cu(s) has E° = 0.78 V and n = 2 at 298 K. At one moment its emf is 0.85 V. What is the reaction quotient Q at that moment? →
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Electrochemistry › Galvanic Cells, Electrode Potential and the Nernst Equation
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