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Byju's Answer
Standard XII
Chemistry
Representation of Cell
Which half-ce...
Question
Which half-cell reactions describes what is going on at the anode, from the diagram ?
A
Z
n
→
Z
n
2
+
+
2
e
−
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B
H
2
→
2
H
+
+
2
e
−
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C
2
C
l
−
→
C
l
2
+
2
e
−
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D
S
O
4
−
→
S
+
2
O
2
+
6
e
−
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E
2
H
+
+
2
e
−
→
H
2
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Solution
The correct option is
A
Z
n
→
Z
n
2
+
+
2
e
−
At anode, zinc is oxidized to Zn(II) ions. Loss of electrons is oxidation.
Z
n
→
Z
n
2
+
+
2
e
−
Suggest Corrections
0
Similar questions
Q.
The standard reduction potentials at 298 K for the following half reactions are given against each
Z
n
2
+
(
a
q
)
+
2
e
⇌
Z
n
(
s
)
−
0.762
C
r
3
+
(
a
q
)
+
2
e
⇌
C
r
(
s
)
−
0.740
2
H
+
(
a
q
)
+
2
e
⇌
H
2
(
g
)
0.000
F
e
3
+
(
a
q
)
+
2
e
⇌
F
e
2
+
(
a
q
)
0.770
Which is the strongest reducing agent?
Q.
In which of the following conditions is the potential for the following half-cell reaction maximum?
2
H
+
+
2
e
−
⟶
H
2
Q.
Find the
△
G
∘
of a galvanic cell with
A
g
and
A
l
.
Standard Reduction Potentials at
25
∘
C
Half-Reaction
E
∘
(
V
)
F
2
+
2
e
−
→
2
F
−
2.87
A
g
2
+
e
−
→
A
g
+
1.99
C
o
3
+
+
e
−
→
C
o
2
+
1.82
H
2
O
2
+
2
H
+
+
2
e
−
→
2
H
2
O
1.78
M
n
O
−
4
+
4
H
+
+
3
e
−
→
M
n
O
2
+
2
H
2
O
1.68
2
e
−
+
2
H
+
+
I
O
−
4
→
I
O
−
3
+
H
2
O
1.60
M
n
O
−
4
+
8
H
+
+
5
e
−
→
M
n
2
+
+
4
H
2
O
1.51
A
u
3
+
+
3
e
−
→
A
u
1.50
C
l
2
+
2
e
−
→
2
C
l
−
1.36
O
2
+
4
H
+
+
4
e
−
→
2
H
2
O
1.23
M
n
O
2
+
4
H
+
+
2
e
−
→
M
n
2
+
+
2
H
2
O
1.21
B
r
2
+
2
e
−
→
2
B
r
−
1.09
N
O
−
3
+
4
H
+
+
3
e
−
→
N
O
+
2
H
2
O
0.96
C
l
O
2
+
e
−
→
C
l
O
2
0.954
2
H
g
2
+
+
2
e
−
→
H
g
2
+
2
0.91
A
g
+
+
e
−
→
A
g
0.80
H
g
2
+
2
+
2
e
−
→
2
H
g
0.80
F
e
3
+
+
e
−
→
F
e
2
+
0.77
O
2
+
2
H
+
+
2
e
−
→
H
2
O
2
0.68
M
n
O
−
4
+
e
−
→
M
n
O
2
−
4
0.56
I
2
+
2
e
−
→
2
I
−
0.54
C
u
+
+
e
−
→
C
u
0.52
O
2
+
2
H
2
O
+
4
e
−
→
4
O
H
−
0.40
C
u
2
+
+
2
e
−
→
C
u
0.34
A
g
C
l
+
e
−
→
A
g
+
C
l
−
0.22
C
u
2
+
+
e
−
→
C
u
+
0.16
2
H
+
+
2
e
−
→
H
2
0.00
F
e
3
+
+
3
e
−
→
F
e
−
0.036
P
b
2
+
+
2
e
−
→
P
b
−
0.13
N
i
2
+
+
2
e
−
→
N
i
−
0.23
C
d
2
+
+
2
e
−
→
C
d
−
0.40
F
e
2
+
+
2
e
−
→
F
e
−
0.44
C
r
3
+
+
e
−
→
C
r
2
+
−
0.50
C
r
3
+
+
3
e
−
→
C
r
−
0.73
Z
n
2
+
+
2
e
−
→
Z
n
−
0.76
2
H
2
O
+
2
e
−
→
H
2
+
2
O
H
−
−
0.83
M
n
2
+
+
2
e
−
→
M
n
−
1.18
A
l
3
+
+
3
e
−
→
A
l
−
1.66
H
2
+
2
e
−
→
2
H
−
−
2.23
M
g
2
+
+
2
e
−
→
M
g
−
2.37
N
a
+
+
e
−
→
N
a
−
2.71
C
a
2
+
+
2
e
−
→
C
a
−
2.76
B
a
2
+
+
2
e
−
→
B
a
−
2.90
K
+
+
e
−
→
K
−
2.92
L
i
+
+
e
−
→
L
i
−
3.05
Q.
Half Reaction
Eo
I.
M
g
→
M
g
2
+
+
2
e
−
2.37V
II.
M
n
→
M
n
2
+
+
2
e
−
1.03V
III.
H
2
→
2
H
+
+
2
e
−
0.00V
IV.
C
u
→
C
u
2
+
+
2
e
−
-0.16V
From the above experimental data what will be the standard reduction potential which represents spontaneous process?
Q.
Rank the following elements by increasing ease of oxidation with most easily oxidized LAST:
C
d
,
C
u
,
N
i
,
F
e
Standard Reduction Potentials at
25
∘
C
Half-Reaction
E
∘
(
V
)
F
2
+
2
e
−
→
2
F
−
2.87
A
g
2
+
e
−
→
A
g
+
1.99
C
o
3
+
+
e
−
→
C
o
2
+
1.82
H
2
O
2
+
2
H
+
+
2
e
−
→
2
H
2
O
1.78
M
n
O
−
4
+
4
H
+
+
3
e
−
→
M
n
O
2
+
2
H
2
O
1.68
2
e
−
+
2
H
+
+
I
O
−
4
→
I
O
−
3
+
H
2
O
1.60
M
n
O
−
4
+
8
H
+
+
5
e
−
→
M
n
2
+
+
4
H
2
O
1.51
A
u
3
+
+
3
e
−
→
A
u
1.50
C
l
2
+
2
e
−
→
2
C
l
−
1.36
O
2
+
4
H
+
+
4
e
−
→
2
H
2
O
1.23
M
n
O
2
+
4
H
+
+
2
e
−
→
M
n
2
+
+
2
H
2
O
1.21
B
r
2
+
2
e
−
→
2
B
r
−
1.09
N
O
−
3
+
4
H
+
+
3
e
−
→
N
O
+
2
H
2
O
0.96
C
l
O
2
+
e
−
→
C
l
O
2
0.954
2
H
g
2
+
+
2
e
−
→
H
g
2
+
2
0.91
A
g
+
+
e
−
→
A
g
0.80
H
g
2
+
2
+
2
e
−
→
2
H
g
0.80
F
e
3
+
+
e
−
→
F
e
2
+
0.77
O
2
+
2
H
+
+
2
e
−
→
H
2
O
2
0.68
M
n
O
−
4
+
e
−
→
M
n
O
2
−
4
0.56
I
2
+
2
e
−
→
2
I
−
0.54
C
u
+
+
e
−
→
C
u
0.52
O
2
+
2
H
2
O
+
4
e
−
→
4
O
H
−
0.40
C
u
2
+
+
2
e
−
→
C
u
0.34
A
g
C
l
+
e
−
→
A
g
+
C
l
−
0.22
C
u
2
+
+
e
−
→
C
u
+
0.16
2
H
+
+
2
e
−
→
H
2
0.00
F
e
3
+
+
3
e
−
→
F
e
−
0.036
P
b
2
+
+
2
e
−
→
P
b
−
0.13
N
i
2
+
+
2
e
−
→
N
i
−
0.23
C
d
2
+
+
2
e
−
→
C
d
−
0.40
F
e
2
+
+
2
e
−
→
F
e
−
0.44
C
r
3
+
+
e
−
→
C
r
2
+
−
0.50
C
r
3
+
+
3
e
−
→
C
r
−
0.73
Z
n
2
+
+
2
e
−
→
Z
n
−
0.76
2
H
2
O
+
2
e
−
→
H
2
+
2
O
H
−
−
0.83
M
n
2
+
+
2
e
−
→
M
n
−
1.18
A
l
3
+
+
3
e
−
→
A
l
−
1.66
H
2
+
2
e
−
→
2
H
−
−
2.23
M
g
2
+
+
2
e
−
→
M
g
−
2.37
N
a
+
+
e
−
→
N
a
−
2.71
C
a
2
+
+
2
e
−
→
C
a
−
2.76
B
a
2
+
+
2
e
−
→
B
a
−
2.90
K
+
+
e
−
→
K
−
2.92
L
i
+
+
e
−
→
L
i
−
3.05
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