Cell Potential Under Nonstandard Conditions
AP Chemistry · Flashcards · Application
AP CHEMISTRY · FLASHCARDS
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QUESTION
Find
Q
Q
Q
at
25
∘
C
25^\circ\text{C}
2
5
∘
C
if
E
∘
=
0.50
V
E^\circ=0.50\ \text{V}
E
∘
=
0.50
V
,
E
=
0.44
V
E=0.44\ \text{V}
E
=
0.44
V
, and
n
=
2
n=2
n
=
2
.
QUESTION
Calculate
E
E
E
at
25
∘
C
25^\circ\text{C}
2
5
∘
C
:
E
∘
=
0.80
V
E^\circ=0.80\ \text{V}
E
∘
=
0.80
V
,
n
=
1
n=1
n
=
1
,
Q
=
10
3
Q=10^{3}
Q
=
1
0
3
.
QUESTION
Calculate
E
E
E
at
25
∘
C
25^\circ\text{C}
2
5
∘
C
:
E
∘
=
1.10
V
E^\circ=1.10\ \text{V}
E
∘
=
1.10
V
,
n
=
2
n=2
n
=
2
,
Q
=
10
−
4
Q=10^{-4}
Q
=
1
0
−
4
.
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ANSWER
1.22
V
1.22\ \text{V}
1.22
V
.
E
=
1.10
−
0.0592
2
log
(
10
−
4
)
=
1.10
+
0.118
=
1.22
E=1.10-\frac{0.0592}{2}\log(10^{-4})=1.10+0.118=1.22
E
=
1.10
−
2
0.0592
lo
g
(
1
0
−
4
)
=
1.10
+
0.118
=
1.22
V.
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