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Electrochemistry: Reduction potentials in a Galvanic Cell
Related Topics
Wize High School Grade 12 Chemistry Textbook > Electrochemistry
Galvanic Cells
7 Activities
Wize High School Grade 12 Chemistry Textbook > Electrochemistry
Reduction Potentials
5 Activities
Use the standard reduction potentials below to answer the following question: which metal could be used to create a sacrificial anode for an aluminum pipe?
R
e
d
u
c
t
i
o
n
h
a
l
f
β
r
e
a
c
t
i
o
n
E
Β°
(
V
)
πΆ
π
2
+
(
π
π
)
+
2
π
β
β
πΆ
π
(
π
)
β
2.87
π
π
+
(
π
π
)
+
π
β
β
π
π
(
π
)
β
2.71
π΄
π
3
+
(
π
π
)
+
3
π
β
β
π΄
π
(
π
)
β
1.66
π
π
2
+
(
π
π
)
+
2
π
β
β
π
π
(
π
)
β
1.18
πΆ
π’
2
+
(
π
π
)
+
2
π
β
β
πΆ
π’
(
π
)
+
0.15
π΄
π
+
(
π
π
)
+
π
β
β
π΄
π
(
π
)
+
0.80
\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ^{2+} (ππ)+2 π^ββπΆπ (π )& -2.87\\ \hline ππ^+ (ππ)+π^ββππ (π )&-2.71\\ \hline π΄π^{3+} (ππ)+3 π^ββπ΄π(π ) &-1.66\\ \hline ππ^{2+} (ππ)+2 π^ββππ (π )&-1.18\\ \hline πΆπ’^{2+} (ππ)+2 π^β βπΆπ’ (π ) &+0.15\\ \hline π΄π^+ (ππ)+π^β βπ΄π (π )&+0.80\end{array}
Reduction
half
β
reaction
C
a
2
+
(
a
q
)
+
2
e
β
β
C
a
(
s
)
N
a
+
(
a
q
)
+
e
β
β
N
a
(
s
)
A
l
3
+
(
a
q
)
+
3
e
β
β
A
l
(
s
)
M
n
2
+
(
a
q
)
+
2
e
β
β
M
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
β
E
Β°
(
V
)
β
2.87
β
2.71
β
1.66
β
1.18
+
0.15
+
0.80
β
β
Ag
Cu
Mn
Ca
I don't know
Check Submission
More Galvanic Cells Questions:
Practice: Galvanic Cells
Galvanic Cell Practice
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
N
a
(
a
q
)
+
+
e
β
β
N
a
(
s
)
E
o
=
β
2.71
V
Na^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
N
a
(
a
q
)
+
β
+
e
β
β
N
a
(
s
)
β
E
o
=
β
2.71
V
Practice: Galvanic Cells
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
N
a
(
a
q
)
+
+
e
β
β
N
a
(
s
)
E
o
=
β
2.71
V
Na^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
N
a
(
a
q
)
+
β
+
e
β
β
N
a
(
s
)
β
E
o
=
β
2.71
V
S
2
O
8
(
a
q
)
2
β
+
2
e
β
β
2
S
O
4
(
a
q
)
2
β
E
o
=
2.01
V
S_2O_{8(aq)}^{2-} +2e^- \to 2SO_{4(aq)}^{2-} \hspace{20pt} E^o=2.01V
S
2
β
O
8
(
a
q
)
2
β
β
+
2
e
β
β
2
S
O
4
(
a
q
)
2
β
β
E
o
=
2.01
V
Practice: Galvanic Cells
Galvanic Cell Practice
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
N
a
(
a
q
)
+
+
e
β
β
N
a
(
s
)
E
o
=
β
2.71
V
Na^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
N
a
(
a
q
)
+
β
+
e
β
β
N
a
(
s
)
β
E
o
=
β
2.71
V
Electrochemistry: Galvanic Cells
In a galvanic cell, oxidation occurs at the:
Electrochemistry: Galvanic Cell
In a zinc-lead cell the reaction is:
P
b
2
+
(
a
q
)
+
Z
n
(
s
)
β
P
b
(
s
)
+
Z
n
2
+
(
a
q
)
E
Β°
=
0.637
V
Pb^{2+}\left(aq\right)+Zn\left(s\right)\to Pb\left(s\right)+Zn^{2+}\left(aq\right)\ \ \ \ \ \ \ E\degree=0.637V
P
b
2
+
(
a
q
)
+
Z
n
(
s
)
β
P
b
(
s
)
+
Z
n
2
+
(
a
q
)
E
Β°
=
0.637
V
Which of the following statements about this cell is FALSE?
Electrochemistry: Galvanic Cells
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
N
a
(
a
q
)
+
+
e
β
β
N
a
(
s
)
E
o
=
β
2.71
V
Na^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
N
a
(
a
q
)
+
β
+
e
β
β
N
a
(
s
)
β
E
o
=
β
2.71
V
S
2
O
8
(
a
q
)
2
β
+
2
e
β
β
2
S
O
4
(
a
q
)
2
β
E
o
=
2.01
V
S_2O_{8(aq)}^{2-} +2e^- \to 2SO_{4(aq)}^{2-} \hspace{20pt} E^o=2.01V
S
2
β
O
8
(
a
q
)
2
β
β
+
2
e
β
β
2
S
O
4
(
a
q
)
2
β
β
E
o
=
2.01
V
A voltaic cell is constructed with an Sn/Sn
2+
half-cell and a Zn/Zn
2+
half-cell. The zinc electrode is negative.
Write a balanced overall reaction for this cell
Draw a diagram of this cell, labeling electrodes with their charges and showing the direction of electron flow in the wire and anion/cation flow in the salt bridge
Choose the INCORRECT statement
Electrochemistry: Standard Reduction Potentials
A galvanic cell is constructed from chromium and copper electrodes immersed in aqueous solutions of Cr
3+
and Cu
2+
respectively (half-reactions shown below).
C
r
3
+
(
a
q
)
+
3
e
β
β
C
r
(
s
)
Cr^{3+}(aq)+3e^-βCr(s)
C
r
3
+
(
a
q
)
+
3
e
β
β
C
r
(
s
)
E
Β°
r
e
d
=
β
0.74
V
EΒ°_{red}=-0.74V
E
Β°
r
e
d
β
=
β
0.74
V
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
Cu^{2+}(aq)+2e^-βCu(s)
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
E
Β°
r
e
d
=
+
0.34
V
EΒ°_{red}=+0.34V
E
Β°
r
e
d
β
=
+
0.34
V
A voltaic cell is constructed with an Sn/Sn
2+
half-cell and a Zn/Zn
2+
half-cell. The zinc electrode is negative.
Draw a diagram of this cell, labeling electrodes with their charges and showing the direction of electron flow in the wire and anion/cation flow in the salt bridge
Electrochemistry: Battery
Given the following half cells, what is the maximum voltage of a battery that could be constructed?
L
i
(
a
q
)
+
+
e
β
β
L
i
(
s
)
E
o
=
β
3.04
V
C
e
(
a
q
)
4
+
+
e
β
β
C
e
(
s
)
3
+
E
o
=
1.44
V
C
o
(
a
q
)
3
+
+
e
β
β
C
o
(
a
q
)
2
+
E
o
=
1.82
V
\begin{array}{llll}&Li^+_{(aq)}&+&e^{-}&\rightarrow&Li_{(s)}&&E^o=-3.04\ V\\ \\&Ce^{4+}_{(aq)}&+&e^{-}&\rightarrow&Ce^{3+}_{(s)}&&E^o=1.44 \ V \\ \\&Co^{3+}_{(aq)}&+&e^{-}&\rightarrow&Co^{2+}_{(aq)}&&E^o=1.82 \ V \end{array}
β
L
i
(
a
q
)
+
β
C
e
(
a
q
)
4
+
β
C
o
(
a
q
)
3
+
β
β
+
+
+
β
e
β
e
β
e
β
β
β
β
β
β
L
i
(
s
)
β
C
e
(
s
)
3
+
β
C
o
(
a
q
)
2
+
β
β
β
E
o
=
β
3.04
V
E
o
=
1.44
V
E
o
=
1.82
V
β
Electrochemistry: Ksp calculations
Use the following reduction potentials to calculate the K
sp
of La(OH)
3(s)
L
a
(
O
H
)
3
(
s
)
+
3
e
β
β
L
a
(
s
)
+
3
O
H
(
a
q
)
β
E
o
=
β
2.90
V
L
a
(
a
q
)
3
+
+
3
e
β
β
L
a
(
s
)
E
o
=
β
2.379
V
\begin{array}{llllll}&La(OH)_{3(s)}&+&3e-&\rightarrow& La_{(s)}&+&3\ OH^-_{(aq)}& &&E^o=-2.90 V \\ &La^{3+}_{(aq)}&+&3e^-&\rightarrow& La_{(s)}&&&&& E^o=-2.379\ V \end{array}
β
L
a
(
O
H
)
3
(
s
)
β
L
a
(
a
q
)
3
+
β
β
+
+
β
3
e
β
3
e
β
β
β
β
β
L
a
(
s
)
β
L
a
(
s
)
β
β
+
β
3
O
H
(
a
q
)
β
β
β
β
β
E
o
=
β
2.90
V
E
o
=
β
2.379
V
β
More Reduction Potentials Questions:
Electrochemistry: Reduction potential application
Reduction Potentials Application
Below we see a diagram showing mitochondrial cellular respiration.
If we just consider the electron carrier NADH, NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen (O2) is the final electron acceptor.
Electrochemistry: Reduction potential application
Reduction Potentials Application
Below we see a diagram showing mitochondrial cellular respiration.
If we just consider the electron carrier NADH, NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen (O2) is the final electron acceptor.
Electrochemistry: Reduction Potentials
Based on the following reduction potentials, which of these species is the strongest reducing agent?
R
e
a
c
t
i
o
n
E
Β°
(
V
)
2
H
C
β
O
(
a
q
)
+
2
H
+
(
a
q
)
+
2
e
β
β
C
β
2
(
g
)
+
2
H
2
O
(
β
)
+
1.63
C
β
2
(
g
)
+
2
e
β
β
2
C
β
β
(
a
q
)
+
1.36
B
r
2
(
β
)
+
2
e
β
β
2
B
r
β
(
a
q
)
+
1.08
I
2
(
s
)
+
2
e
β
β
2
I
β
(
a
q
)
+
0.535
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
+
0.799
S
n
4
+
(
a
q
)
+
2
e
β
β
S
n
2
+
(
a
q
)
+
0.15
\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & EΒ° (V) \\ \hline 2 HCβO (aq) + 2 H^+ (aq) + 2 e^β β Cβ_2 (g) + 2 H_2O (β) & + 1.63 \\ \hline Cβ_2(g) + 2 e^β β 2 Cβ^β (aq) &+ 1.36\\ \hline Br_2(β) + 2 e^β β 2 Br^β (aq) & + 1.08\\ \hline I_2 (s) + 2 e^β β 2 I^β (aq) & + 0.535\\ \hline Ag+ (aq) + e^β β Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^β β Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}
Reaction
2
H
C
β
O
(
a
q
)
+
2
H
+
(
a
q
)
+
2
e
β
β
C
β
2
β
(
g
)
+
2
H
2
β
O
(
β
)
C
β
2
β
(
g
)
+
2
e
β
β
2
C
β
β
(
a
q
)
B
r
2
β
(
β
)
+
2
e
β
β
2
B
r
β
(
a
q
)
I
2
β
(
s
)
+
2
e
β
β
2
I
β
(
a
q
)
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
S
n
4
+
(
a
q
)
+
2
e
β
β
S
n
2
+
(
a
q
)
β
E
Β°
(
V
)
+
1.63
+
1.36
+
1.08
+
0.535
+
0.799
+
0.15
β
β
Electrochemistry: Reduction potential application
Reduction Potentials Application
Below we see a diagram showing mitochondrial cellular respiration.
If we just consider the electron carrier NADH, NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen (O2) is the final electron acceptor.
Electrochemical Equilibrium and Cells
Will the following reaction proceed spontaneously or will current need to be added externally?
C
u
(
s
)
+
C
a
(
a
q
)
2
+
β
C
a
(
s
)
+
C
u
(
a
q
)
2
+
Cu_{(s)}+Ca^{2+}_{(aq)} \to Ca_{(s)} +Cu^{2+} _{(aq)}
C
u
(
s
)
β
+
C
a
(
a
q
)
2
+
β
β
C
a
(
s
)
β
+
C
u
(
a
q
)
2
+
β
In basic solution Se
2-
and SO
3
2-
ions react spontaneously:
2
S
e
(
a
q
)
2
β
+
2
S
O
3
(
a
q
)
2
β
+
3
H
2
O
(
l
)
β
2
S
e
(
s
)
+
6
O
H
(
a
q
)
β
+
S
2
O
(
a
q
)
2
β
,
E
c
e
l
l
o
=
0.35
V
2Se_{(aq)}^{2-}+2SO_{3(aq)}^{2-}+3H_2O_{(l)} \to 2Se_{(s)}+6OH_{(aq)}^-+S_2O^{2-}_{(aq)}, E^o_{cell}=0.35V
2
S
e
(
a
q
)
2
β
β
+
2
S
O
3
(
a
q
)
2
β
β
+
3
H
2
β
O
(
l
)
β
β
2
S
e
(
s
)
β
+
6
O
H
(
a
q
)
β
β
+
S
2
β
O
(
a
q
)
2
β
β
,
E
ce
l
l
o
β
=
0.35
V
Write balanced half reactions for this process
(Duplicated)
Use the standard reduction potentials below to answer the following question: which species listed is the strongest oxidizing agent?
R
e
d
u
c
t
i
o
n
h
a
l
f
β
r
e
a
c
t
i
o
n
E
Β°
(
V
)
πΆ
π
2
+
(
π
π
)
+
2
π
β
β
πΆ
π
(
π
)
β
2.87
π
π
+
(
π
π
)
+
π
β
β
π
π
(
π
)
β
2.71
π΄
π
3
+
(
π
π
)
+
3
π
β
β
π΄
π
(
π
)
β
1.66
π
π
2
+
(
π
π
)
+
2
π
β
β
π
π
(
π
)
β
1.18
πΆ
π’
2
+
(
π
π
)
+
2
π
β
β
πΆ
π’
(
π
)
+
0.15
π΄
π
+
(
π
π
)
+
π
β
β
π΄
π
(
π
)
+
0.80
\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ^{2+} (ππ)+2 π^ββπΆπ (π )& -2.87\\ \hline ππ^+ (ππ)+π^ββππ (π )&-2.71\\ \hline π΄π^{3+} (ππ)+3 π^ββπ΄π(π ) &-1.66\\ \hline ππ^{2+} (ππ)+2 π^ββππ (π )&-1.18\\ \hline πΆπ’^{2+} (ππ)+2 π^β βπΆπ’ (π ) &+0.15\\ \hline π΄π^+ (ππ)+π^β βπ΄π (π )&+0.80\end{array}
Reduction
half
β
reaction
C
a
2
+
(
a
q
)
+
2
e
β
β
C
a
(
s
)
N
a
+
(
a
q
)
+
e
β
β
N
a
(
s
)
A
l
3
+
(
a
q
)
+
3
e
β
β
A
l
(
s
)
M
n
2
+
(
a
q
)
+
2
e
β
β
M
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
β
E
Β°
(
V
)
β
2.87
β
2.71
β
1.66
β
1.18
+
0.15
+
0.80
β
β
Electrochemistry: Standard Reduction Potentials
A galvanic cell is constructed from chromium and copper electrodes immersed in aqueous solutions of Cr
3+
and Cu
2+
respectively (half-reactions shown below).
C
r
3
+
(
a
q
)
+
3
e
β
β
C
r
(
s
)
Cr^{3+}(aq)+3e^-βCr(s)
C
r
3
+
(
a
q
)
+
3
e
β
β
C
r
(
s
)
E
Β°
r
e
d
=
β
0.74
V
EΒ°_{red}=-0.74V
E
Β°
r
e
d
β
=
β
0.74
V
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
Cu^{2+}(aq)+2e^-βCu(s)
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
E
Β°
r
e
d
=
+
0.34
V
EΒ°_{red}=+0.34V
E
Β°
r
e
d
β
=
+
0.34
V
Electrochemistry: Reduction potentials
Based on the data table, which of these species is the strongest reducing agent?
Based on the following reduction potentials, which of these species is the strongest reducing agent?
R
e
a
c
t
i
o
n
E
Β°
(
V
)
2
H
C
β
O
(
a
q
)
+
2
H
+
(
a
q
)
+
2
e
β
β
C
β
2
(
g
)
+
2
H
2
O
(
β
)
+
1.63
C
β
2
(
g
)
+
2
e
β
β
2
C
β
β
(
a
q
)
+
1.36
B
r
2
(
β
)
+
2
e
β
β
2
B
r
β
(
a
q
)
+
1.08
I
2
(
s
)
+
2
e
β
β
2
I
β
(
a
q
)
+
0.535
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
+
0.799
S
n
4
+
(
a
q
)
+
2
e
β
β
S
n
2
+
(
a
q
)
+
0.15
\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & EΒ° (V) \\ \hline 2 HCβO (aq) + 2 H^+ (aq) + 2 e^β β Cβ_2 (g) + 2 H_2O (β) & + 1.63 \\ \hline Cβ_2(g) + 2 e^β β 2 Cβ^β (aq) &+ 1.36\\ \hline Br_2(β) + 2 e^β β 2 Br^β (aq) & + 1.08\\ \hline I_2 (s) + 2 e^β β 2 I^β (aq) & + 0.535\\ \hline Ag+ (aq) + e^β β Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^β β Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}
Reaction
2
H
C
β
O
(
a
q
)
+
2
H
+
(
a
q
)
+
2
e
β
β
C
β
2
β
(
g
)
+
2
H
2
β
O
(
β
)
C
β
2
β
(
g
)
+
2
e
β
β
2
C
β
β
(
a
q
)
B
r
2
β
(
β
)
+
2
e
β
β
2
B
r
β
(
a
q
)
I
2
β
(
s
)
+
2
e
β
β
2
I
β
(
a
q
)
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
S
n
4
+
(
a
q
)
+
2
e
β
β
S
n
2
+
(
a
q
)
β
E
Β°
(
V
)
+
1.63
+
1.36
+
1.08
+
0.535
+
0.799
+
0.15
β
β
Electrochemistry: Oxidizing agents
Which one of the species above is the best oxidizing agent?
Z
n
2
+
(
a
q
)
+
2
e
β
β
Z
n
(
s
)
E
o
=
β
0.74
V
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
E
o
=
+
0.34
V
C
o
2
+
(
a
q
)
+
2
e
β
β
C
o
(
s
)
E
o
=
β
0.28
V
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
E
o
=
+
0.80
V
\def\arraystretch{1.5}\begin{array}{cc}Zn^{2+}(aq)+2e^-βZn(s)&&E^o=-0.74V\\ Cu^{2+}(aq)+2e^-βCu(s)&&E^o=+0.34V\\ Co^{2+}(aq)+2e^-βCo(s)&&E^o=-0.28V\\ Ag^+(aq)+e^-βAg(s)&&E^o=+0.80V\end{array}
Z
n
2
+
(
a
q
)
+
2
e
β
β
Z
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
C
o
2
+
(
a
q
)
+
2
e
β
β
C
o
(
s
)
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
β
β
E
o
=
β
0.74
V
E
o
=
+
0.34
V
E
o
=
β
0.28
V
E
o
=
+
0.80
V
β
Electrochemistry: Galvanic Cells
The cell pictured below is prepared under standard conditions. In this cell, what will be the
anode
?
πΆ
π’
2
+
(
π
π
)
+
π»
2
(
π
)
β
πΆ
π’
(
π
)
+
2
π»
+
(
π
π
)
πΈ
Β°
=
0.339
π
πΆπ’^{2+} (ππ)+π»_2 (π)\rightleftharpoons πΆπ’ (π )+2 π»^+ (ππ) \qquadπΈΒ°=0.339 π
C
u
2
+
(
a
q
)
+
H
2
β
(
g
)
β
C
u
(
s
)
+
2
H
+
(
a
q
)
E
Β°
=
0.339
V
Use the standard reduction potentials below to answer the following question: which metal could be used as an anode for an aluminum pipe?
R
e
d
u
c
t
i
o
n
h
a
l
f
β
r
e
a
c
t
i
o
n
E
Β°
(
V
)
πΆ
π
2
+
(
π
π
)
+
2
π
β
β
πΆ
π
(
π
)
β
2.87
π
π
+
(
π
π
)
+
π
β
β
π
π
(
π
)
β
2.71
π΄
π
3
+
(
π
π
)
+
3
π
β
β
π΄
π
(
π
)
β
1.66
π
π
2
+
(
π
π
)
+
2
π
β
β
π
π
(
π
)
β
1.18
πΆ
π’
2
+
(
π
π
)
+
2
π
β
β
πΆ
π’
(
π
)
+
0.15
π΄
π
+
(
π
π
)
+
π
β
β
π΄
π
(
π
)
+
0.80
\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ^{2+} (ππ)+2 π^ββπΆπ (π )& -2.87\\ \hline ππ^+ (ππ)+π^ββππ (π )&-2.71\\ \hline π΄π^{3+} (ππ)+3 π^ββπ΄π(π ) &-1.66\\ \hline ππ^{2+} (ππ)+2 π^ββππ (π )&-1.18\\ \hline πΆπ’^{2+} (ππ)+2 π^β βπΆπ’ (π ) &+0.15\\ \hline π΄π^+ (ππ)+π^β βπ΄π (π )&+0.80\end{array}
Reduction
half
β
reaction
C
a
2
+
(
a
q
)
+
2
e
β
β
C
a
(
s
)
N
a
+
(
a
q
)
+
e
β
β
N
a
(
s
)
A
l
3
+
(
a
q
)
+
3
e
β
β
A
l
(
s
)
M
n
2
+
(
a
q
)
+
2
e
β
β
M
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
β
E
Β°
(
V
)
β
2.87
β
2.71
β
1.66
β
1.18
+
0.15
+
0.80
β
β
Which one of the species above is the best oxidizing agent?
Z
n
2
+
(
a
q
)
+
2
e
β
β
Z
n
(
s
)
E
o
=
β
0.74
V
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
E
o
=
+
0.34
V
C
o
2
+
(
a
q
)
+
2
e
β
β
C
o
(
s
)
E
o
=
β
0.28
V
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
E
o
=
+
0.80
V
\def\arraystretch{1.5}\begin{array}{cc}Zn^{2+}(aq)+2e^-βZn(s)&&E^o=-0.74V\\ Cu^{2+}(aq)+2e^-βCu(s)&&E^o=+0.34V\\ Co^{2+}(aq)+2e^-βCo(s)&&E^o=-0.28V\\ Ag^+(aq)+e^-βAg(s)&&E^o=+0.80V\end{array}
Z
n
2
+
(
a
q
)
+
2
e
β
β
Z
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
β
β
C
u
(
s
)
C
o
2
+
(
a
q
)
+
2
e
β
β
C
o
(
s
)
A
g
+
(
a
q
)
+
e
β
β
A
g
(
s
)
β
β
E
o
=
β
0.74
V
E
o
=
+
0.34
V
E
o
=
β
0.28
V
E
o
=
+
0.80
V
β
The cell pictured is prepared at standard conditions and the cell voltage measured to be +0.339 V. Which of the following changes would result in an
increase
to the measured cell voltage?
πΆ
π’
2
+
(
π
π
)
+
π»
2
(
π
)
β
πΆ
π’
(
π
)
+
2
π»
+
(
π
π
)
πΈ
Β°
=
0.339
π
πΆπ’^{2+} (ππ)+π»_2 (π)\rightleftharpoons πΆπ’ (π )+2 π»^+ (ππ) \qquadπΈΒ°=0.339 π
C
u
2
+
(
a
q
)
+
H
2
β
(
g
)
β
C
u
(
s
)
+
2
H
+
(
a
q
)
E
Β°
=
0.339
V