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Electrochemistry: Oxidizing agents
Related Topics
Wize High School Grade 12 Chemistry Textbook > Electrochemistry
Reduction Potentials
5 Activities
Which one of the species above is the best oxidizing agent?
Z
n
2
+
(
a
q
)
+
2
e
−
⇌
Z
n
(
s
)
E
o
=
−
0.74
V
C
u
2
+
(
a
q
)
+
2
e
−
⇌
C
u
(
s
)
E
o
=
+
0.34
V
C
o
2
+
(
a
q
)
+
2
e
−
⇌
C
o
(
s
)
E
o
=
−
0.28
V
A
g
+
(
a
q
)
+
e
−
⇌
A
g
(
s
)
E
o
=
+
0.80
V
\def\arraystretch{1.5}\begin{array}{cc}Zn^{2+}(aq)+2e^-⇌Zn(s)&&E^o=-0.74V\\ Cu^{2+}(aq)+2e^-⇌Cu(s)&&E^o=+0.34V\\ Co^{2+}(aq)+2e^-⇌Co(s)&&E^o=-0.28V\\ Ag^+(aq)+e^-⇌Ag(s)&&E^o=+0.80V\end{array}
Z
n
2
+
(
a
q
)
+
2
e
−
⇌
Z
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
−
⇌
C
u
(
s
)
C
o
2
+
(
a
q
)
+
2
e
−
⇌
C
o
(
s
)
A
g
+
(
a
q
)
+
e
−
⇌
A
g
(
s
)
E
o
=
−
0.74
V
E
o
=
+
0.34
V
E
o
=
−
0.28
V
E
o
=
+
0.80
V
C
o
(
s
)
Co(s)
C
o
(
s
)
Z
n
(
s
)
Zn(s)
Z
n
(
s
)
C
u
2
+
(
a
q
)
Cu^{2+}(aq)
C
u
2
+
(
a
q
)
C
u
(
s
)
Cu(s)
C
u
(
s
)
A
g
+
(
a
q
)
Ag^+(aq)
A
g
+
(
a
q
)
I don't know
Check Submission
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
°
(
V
)
2
H
C
ℓ
O
(
a
q
)
+
2
H
+
(
a
q
)
+
2
e
−
→
C
ℓ
2
(
g
)
+
2
H
2
O
(
ℓ
)
+
1.63
C
ℓ
2
(
g
)
+
2
e
−
→
2
C
ℓ
−
(
a
q
)
+
1.36
B
r
2
(
ℓ
)
+
2
e
−
→
2
B
r
−
(
a
q
)
+
1.08
I
2
(
s
)
+
2
e
−
→
2
I
−
(
a
q
)
+
0.535
A
g
+
(
a
q
)
+
e
−
→
A
g
(
s
)
+
0.799
S
n
4
+
(
a
q
)
+
2
e
−
→
S
n
2
+
(
a
q
)
+
0.15
\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & E° (V) \\ \hline 2 HCℓO (aq) + 2 H^+ (aq) + 2 e^− → Cℓ_2 (g) + 2 H_2O (ℓ) & + 1.63 \\ \hline Cℓ_2(g) + 2 e^− → 2 Cℓ^− (aq) &+ 1.36\\ \hline Br_2(ℓ) + 2 e^− → 2 Br^− (aq) & + 1.08\\ \hline I_2 (s) + 2 e^− → 2 I^− (aq) & + 0.535\\ \hline Ag+ (aq) + e^− → Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^− → Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}
Reaction
2
H
C
ℓ
O
(
a
q
)
+
2
H
+
(
a
q
)
+
2
e
−
→
C
ℓ
2
(
g
)
+
2
H
2
O
(
ℓ
)
C
ℓ
2
(
g
)
+
2
e
−
→
2
C
ℓ
−
(
a
q
)
B
r
2
(
ℓ
)
+
2
e
−
→
2
B
r
−
(
a
q
)
I
2
(
s
)
+
2
e
−
→
2
I
−
(
a
q
)
A
g
+
(
a
q
)
+
e
−
→
A
g
(
s
)
S
n
4
+
(
a
q
)
+
2
e
−
→
S
n
2
+
(
a
q
)
E
°
(
V
)
+
1.63
+
1.36
+
1.08
+
0.535
+
0.799
+
0.15
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
+
→
C
a
(
s
)
+
C
u
(
a
q
)
2
+
Cu_{(s)}+Ca^{2+}_{(aq)} \to Ca_{(s)} +Cu^{2+} _{(aq)}
C
u
(
s
)
+
C
a
(
a
q
)
2
+
→
C
a
(
s
)
+
C
u
(
a
q
)
2
+
In basic solution Se
2-
and SO
3
2-
ions react spontaneously:
2
S
e
(
a
q
)
2
−
+
2
S
O
3
(
a
q
)
2
−
+
3
H
2
O
(
l
)
→
2
S
e
(
s
)
+
6
O
H
(
a
q
)
−
+
S
2
O
(
a
q
)
2
−
,
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
−
+
2
S
O
3
(
a
q
)
2
−
+
3
H
2
O
(
l
)
→
2
S
e
(
s
)
+
6
O
H
(
a
q
)
−
+
S
2
O
(
a
q
)
2
−
,
E
ce
l
l
o
=
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
−
r
e
a
c
t
i
o
n
E
°
(
V
)
𝐶
𝑎
2
+
(
𝑎
𝑞
)
+
2
𝑒
−
⇌
𝐶
𝑎
(
𝑠
)
−
2.87
𝑁
𝑎
+
(
𝑎
𝑞
)
+
𝑒
−
⇌
𝑁
𝑎
(
𝑠
)
−
2.71
𝐴
𝑙
3
+
(
𝑎
𝑞
)
+
3
𝑒
−
⇌
𝐴
𝑙
(
𝑠
)
−
1.66
𝑀
𝑛
2
+
(
𝑎
𝑞
)
+
2
𝑒
−
⇌
𝑀
𝑛
(
𝑠
)
−
1.18
𝐶
𝑢
2
+
(
𝑎
𝑞
)
+
2
𝑒
−
⇌
𝐶
𝑢
(
𝑠
)
+
0.15
𝐴
𝑔
+
(
𝑎
𝑞
)
+
𝑒
−
⇌
𝐴
𝑔
(
𝑠
)
+
0.80
\begin{array}{c:c}\rm Reduction\ half-reaction& E° (V)\\ \hline 𝐶𝑎^{2+} (𝑎𝑞)+2 𝑒^−⇌𝐶𝑎 (𝑠)& -2.87\\ \hline 𝑁𝑎^+ (𝑎𝑞)+𝑒^−⇌𝑁𝑎 (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (𝑎𝑞)+3 𝑒^−⇌𝐴𝑙(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (𝑎𝑞)+2 𝑒^−⇌𝑀𝑛 (𝑠)&-1.18\\ \hline 𝐶𝑢^{2+} (𝑎𝑞)+2 𝑒^− ⇌𝐶𝑢 (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (𝑎𝑞)+𝑒^− ⇌𝐴𝑔 (𝑠)&+0.80\end{array}
Reduction
half
−
reaction
C
a
2
+
(
a
q
)
+
2
e
−
⇌
C
a
(
s
)
N
a
+
(
a
q
)
+
e
−
⇌
N
a
(
s
)
A
l
3
+
(
a
q
)
+
3
e
−
⇌
A
l
(
s
)
M
n
2
+
(
a
q
)
+
2
e
−
⇌
M
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
−
⇌
C
u
(
s
)
A
g
+
(
a
q
)
+
e
−
⇌
A
g
(
s
)
E
°
(
V
)
−
2.87
−
2.71
−
1.66
−
1.18
+
0.15
+
0.80
Electrochemistry: Reduction potentials in a Galvanic Cell
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
−
r
e
a
c
t
i
o
n
E
°
(
V
)
𝐶
𝑎
2
+
(
𝑎
𝑞
)
+
2
𝑒
−
⇌
𝐶
𝑎
(
𝑠
)
−
2.87
𝑁
𝑎
+
(
𝑎
𝑞
)
+
𝑒
−
⇌
𝑁
𝑎
(
𝑠
)
−
2.71
𝐴
𝑙
3
+
(
𝑎
𝑞
)
+
3
𝑒
−
⇌
𝐴
𝑙
(
𝑠
)
−
1.66
𝑀
𝑛
2
+
(
𝑎
𝑞
)
+
2
𝑒
−
⇌
𝑀
𝑛
(
𝑠
)
−
1.18
𝐶
𝑢
2
+
(
𝑎
𝑞
)
+
2
𝑒
−
⇌
𝐶
𝑢
(
𝑠
)
+
0.15
𝐴
𝑔
+
(
𝑎
𝑞
)
+
𝑒
−
⇌
𝐴
𝑔
(
𝑠
)
+
0.80
\begin{array}{c:c}\rm Reduction\ half-reaction& E° (V)\\ \hline 𝐶𝑎^{2+} (𝑎𝑞)+2 𝑒^−⇌𝐶𝑎 (𝑠)& -2.87\\ \hline 𝑁𝑎^+ (𝑎𝑞)+𝑒^−⇌𝑁𝑎 (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (𝑎𝑞)+3 𝑒^−⇌𝐴𝑙(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (𝑎𝑞)+2 𝑒^−⇌𝑀𝑛 (𝑠)&-1.18\\ \hline 𝐶𝑢^{2+} (𝑎𝑞)+2 𝑒^− ⇌𝐶𝑢 (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (𝑎𝑞)+𝑒^− ⇌𝐴𝑔 (𝑠)&+0.80\end{array}
Reduction
half
−
reaction
C
a
2
+
(
a
q
)
+
2
e
−
⇌
C
a
(
s
)
N
a
+
(
a
q
)
+
e
−
⇌
N
a
(
s
)
A
l
3
+
(
a
q
)
+
3
e
−
⇌
A
l
(
s
)
M
n
2
+
(
a
q
)
+
2
e
−
⇌
M
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
−
⇌
C
u
(
s
)
A
g
+
(
a
q
)
+
e
−
⇌
A
g
(
s
)
E
°
(
V
)
−
2.87
−
2.71
−
1.66
−
1.18
+
0.15
+
0.80
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
−
⇌
C
r
(
s
)
Cr^{3+}(aq)+3e^-⇌Cr(s)
C
r
3
+
(
a
q
)
+
3
e
−
⇌
C
r
(
s
)
E
°
r
e
d
=
−
0.74
V
E°_{red}=-0.74V
E
°
r
e
d
=
−
0.74
V
C
u
2
+
(
a
q
)
+
2
e
−
⇌
C
u
(
s
)
Cu^{2+}(aq)+2e^-⇌Cu(s)
C
u
2
+
(
a
q
)
+
2
e
−
⇌
C
u
(
s
)
E
°
r
e
d
=
+
0.34
V
E°_{red}=+0.34V
E
°
r
e
d
=
+
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
°
(
V
)
2
H
C
ℓ
O
(
a
q
)
+
2
H
+
(
a
q
)
+
2
e
−
→
C
ℓ
2
(
g
)
+
2
H
2
O
(
ℓ
)
+
1.63
C
ℓ
2
(
g
)
+
2
e
−
→
2
C
ℓ
−
(
a
q
)
+
1.36
B
r
2
(
ℓ
)
+
2
e
−
→
2
B
r
−
(
a
q
)
+
1.08
I
2
(
s
)
+
2
e
−
→
2
I
−
(
a
q
)
+
0.535
A
g
+
(
a
q
)
+
e
−
→
A
g
(
s
)
+
0.799
S
n
4
+
(
a
q
)
+
2
e
−
→
S
n
2
+
(
a
q
)
+
0.15
\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & E° (V) \\ \hline 2 HCℓO (aq) + 2 H^+ (aq) + 2 e^− → Cℓ_2 (g) + 2 H_2O (ℓ) & + 1.63 \\ \hline Cℓ_2(g) + 2 e^− → 2 Cℓ^− (aq) &+ 1.36\\ \hline Br_2(ℓ) + 2 e^− → 2 Br^− (aq) & + 1.08\\ \hline I_2 (s) + 2 e^− → 2 I^− (aq) & + 0.535\\ \hline Ag+ (aq) + e^− → Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^− → Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}
Reaction
2
H
C
ℓ
O
(
a
q
)
+
2
H
+
(
a
q
)
+
2
e
−
→
C
ℓ
2
(
g
)
+
2
H
2
O
(
ℓ
)
C
ℓ
2
(
g
)
+
2
e
−
→
2
C
ℓ
−
(
a
q
)
B
r
2
(
ℓ
)
+
2
e
−
→
2
B
r
−
(
a
q
)
I
2
(
s
)
+
2
e
−
→
2
I
−
(
a
q
)
A
g
+
(
a
q
)
+
e
−
→
A
g
(
s
)
S
n
4
+
(
a
q
)
+
2
e
−
→
S
n
2
+
(
a
q
)
E
°
(
V
)
+
1.63
+
1.36
+
1.08
+
0.535
+
0.799
+
0.15
Electrochemistry: Galvanic Cells
The cell pictured below is prepared under standard conditions. In this cell, what will be the
anode
?
𝐶
𝑢
2
+
(
𝑎
𝑞
)
+
𝐻
2
(
𝑔
)
⇌
𝐶
𝑢
(
𝑠
)
+
2
𝐻
+
(
𝑎
𝑞
)
𝐸
°
=
0.339
𝑉
𝐶𝑢^{2+} (𝑎𝑞)+𝐻_2 (𝑔)\rightleftharpoons 𝐶𝑢 (𝑠)+2 𝐻^+ (𝑎𝑞) \qquad𝐸°=0.339 𝑉
C
u
2
+
(
a
q
)
+
H
2
(
g
)
⇌
C
u
(
s
)
+
2
H
+
(
a
q
)
E
°
=
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
−
r
e
a
c
t
i
o
n
E
°
(
V
)
𝐶
𝑎
2
+
(
𝑎
𝑞
)
+
2
𝑒
−
⇌
𝐶
𝑎
(
𝑠
)
−
2.87
𝑁
𝑎
+
(
𝑎
𝑞
)
+
𝑒
−
⇌
𝑁
𝑎
(
𝑠
)
−
2.71
𝐴
𝑙
3
+
(
𝑎
𝑞
)
+
3
𝑒
−
⇌
𝐴
𝑙
(
𝑠
)
−
1.66
𝑀
𝑛
2
+
(
𝑎
𝑞
)
+
2
𝑒
−
⇌
𝑀
𝑛
(
𝑠
)
−
1.18
𝐶
𝑢
2
+
(
𝑎
𝑞
)
+
2
𝑒
−
⇌
𝐶
𝑢
(
𝑠
)
+
0.15
𝐴
𝑔
+
(
𝑎
𝑞
)
+
𝑒
−
⇌
𝐴
𝑔
(
𝑠
)
+
0.80
\begin{array}{c:c}\rm Reduction\ half-reaction& E° (V)\\ \hline 𝐶𝑎^{2+} (𝑎𝑞)+2 𝑒^−⇌𝐶𝑎 (𝑠)& -2.87\\ \hline 𝑁𝑎^+ (𝑎𝑞)+𝑒^−⇌𝑁𝑎 (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (𝑎𝑞)+3 𝑒^−⇌𝐴𝑙(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (𝑎𝑞)+2 𝑒^−⇌𝑀𝑛 (𝑠)&-1.18\\ \hline 𝐶𝑢^{2+} (𝑎𝑞)+2 𝑒^− ⇌𝐶𝑢 (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (𝑎𝑞)+𝑒^− ⇌𝐴𝑔 (𝑠)&+0.80\end{array}
Reduction
half
−
reaction
C
a
2
+
(
a
q
)
+
2
e
−
⇌
C
a
(
s
)
N
a
+
(
a
q
)
+
e
−
⇌
N
a
(
s
)
A
l
3
+
(
a
q
)
+
3
e
−
⇌
A
l
(
s
)
M
n
2
+
(
a
q
)
+
2
e
−
⇌
M
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
−
⇌
C
u
(
s
)
A
g
+
(
a
q
)
+
e
−
⇌
A
g
(
s
)
E
°
(
V
)
−
2.87
−
2.71
−
1.66
−
1.18
+
0.15
+
0.80
Which one of the species above is the best oxidizing agent?
Z
n
2
+
(
a
q
)
+
2
e
−
⇌
Z
n
(
s
)
E
o
=
−
0.74
V
C
u
2
+
(
a
q
)
+
2
e
−
⇌
C
u
(
s
)
E
o
=
+
0.34
V
C
o
2
+
(
a
q
)
+
2
e
−
⇌
C
o
(
s
)
E
o
=
−
0.28
V
A
g
+
(
a
q
)
+
e
−
⇌
A
g
(
s
)
E
o
=
+
0.80
V
\def\arraystretch{1.5}\begin{array}{cc}Zn^{2+}(aq)+2e^-⇌Zn(s)&&E^o=-0.74V\\ Cu^{2+}(aq)+2e^-⇌Cu(s)&&E^o=+0.34V\\ Co^{2+}(aq)+2e^-⇌Co(s)&&E^o=-0.28V\\ Ag^+(aq)+e^-⇌Ag(s)&&E^o=+0.80V\end{array}
Z
n
2
+
(
a
q
)
+
2
e
−
⇌
Z
n
(
s
)
C
u
2
+
(
a
q
)
+
2
e
−
⇌
C
u
(
s
)
C
o
2
+
(
a
q
)
+
2
e
−
⇌
C
o
(
s
)
A
g
+
(
a
q
)
+
e
−
⇌
A
g
(
s
)
E
o
=
−
0.74
V
E
o
=
+
0.34
V
E
o
=
−
0.28
V
E
o
=
+
0.80
V
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?
𝐶
𝑢
2
+
(
𝑎
𝑞
)
+
𝐻
2
(
𝑔
)
⇌
𝐶
𝑢
(
𝑠
)
+
2
𝐻
+
(
𝑎
𝑞
)
𝐸
°
=
0.339
𝑉
𝐶𝑢^{2+} (𝑎𝑞)+𝐻_2 (𝑔)\rightleftharpoons 𝐶𝑢 (𝑠)+2 𝐻^+ (𝑎𝑞) \qquad𝐸°=0.339 𝑉
C
u
2
+
(
a
q
)
+
H
2
(
g
)
⇌
C
u
(
s
)
+
2
H
+
(
a
q
)
E
°
=
0.339
V