Electrochemical Cells and Redox Reactions (4C)
MCAT · Flashcards · Application
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QUESTION
At
298
K
298\,\text{K}
298
K
, find
E
E
E
if
E
∘
=
0.30
V
E^\circ=0.30\,\text{V}
E
∘
=
0.30
V
,
n
=
2
n=2
n
=
2
, and
Q
=
10
Q=10
Q
=
10
.
QUESTION
At
298
K
298\,\text{K}
298
K
, find
log
K
\log K
lo
g
K
if
n
=
1
n=1
n
=
1
and
E
cell
∘
=
0.0592
V
E^\circ_{\text{cell}}=0.0592\,\text{V}
E
cell
∘
=
0.0592
V
.
QUESTION
Find
E
cell
∘
E^\circ_{\text{cell}}
E
cell
∘
if
E
cathode
∘
=
+
0.80
V
E^\circ_{\text{cathode}}=+0.80\,\text{V}
E
cathode
∘
=
+
0.80
V
and
E
anode
∘
=
−
0.20
V
E^\circ_{\text{anode}}=-0.20\,\text{V}
E
anode
∘
=
−
0.20
V
(both reduction potentials).
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ANSWER
E
cell
∘
=
+
1.00
V
E^\circ_{\text{cell}}=+1.00\,\text{V}
E
cell
∘
=
+
1.00
V
. Using
E
cell
∘
=
E
cathode
∘
−
E
anode
∘
E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}
E
cell
∘
=
E
cathode
∘
−
E
anode
∘
, substitute values to get
+
0.80
−
(
−
0.20
)
=
+
1.00
V
+0.80 - (-0.20)=+1.00\,\text{V}
+
0.80
−
(
−
0.20
)
=
+
1.00
V
.
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