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Byju's Answer
Standard IX
Chemistry
Electrolysis of Water
During the el...
Question
During the electrolysis of dilute sulphuric acid, the following process is possible at anode.
A
2
H
2
O
(
l
)
→
O
2
(
g
)
+
4
H
+
(
a
q
)
+
4
e
−
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B
2
S
O
2
−
4
(
a
q
)
→
S
2
O
2
−
S
(
a
q
)
+
2
e
−
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C
H
2
O
(
l
)
→
H
+
(
a
q
)
+
O
H
−
(
a
q
)
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D
H
2
O
(
l
)
+
e
−
→
1
2
H
2
(
g
)
+
O
H
−
(
a
q
)
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Solution
The correct option is
A
2
H
2
O
(
l
)
→
O
2
(
g
)
+
4
H
+
(
a
q
)
+
4
e
−
During electrolysis of
H
2
S
O
4
H
2
is evolved at cathode.
2
H
+
+
2
e
−
⟶
H
2
O
2
is evolved at anode,
2
H
2
O
⟶
O
2
+
4
H
+
+
4
e
−
Suggest Corrections
1
Similar questions
Q.
During electrolysis,
O
2
(
g
)
is evolved at anode in
⎡
⎢ ⎢ ⎢ ⎢ ⎢
⎣
(
c
o
r
r
e
c
t
)
C
u
(
s
)
⟶
C
u
2
+
(
a
q
)
+
2
e
−
E
⊝
C
u
|
C
u
2
+
=
−
0.34
V
(
w
r
o
n
g
)
2
H
2
O
(
l
)
⟶
O
2
(
g
)
+
4
H
⊕
(
a
q
)
+
4
e
−
E
⊝
H
2
O
|
O
2
=
−
1.23
V
⎤
⎥ ⎥ ⎥ ⎥ ⎥
⎦
I
n
(
d
)
⎡
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢
⎣
(
c
o
r
r
e
c
t
)
N
i
(
s
)
⟶
N
i
2
+
(
a
q
)
+
2
e
−
E
⊝
N
i
|
N
i
2
+
=
−
0.25
V
(
w
r
o
n
g
)
4
⊝
O
H
(
a
q
)
⟶
O
2
(
g
)
+
2
H
2
O
(
l
)
+
4
e
−
E
⊕
⊝
O
H
|
O
2
=
−
0.4
V
⎤
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
⎦
Q.
Match the column I with column II and mark the appropriate choice.
Column I
Column II
(A)
P
b
(
s
)
+
S
O
−
2
4
(
a
q
)
→
P
b
S
O
4
(
s
)
+
2
e
−
(i)
Rusting of iron
(B)
2
S
O
2
−
4
(
a
q
)
→
S
2
O
2
−
8
(
a
q
)
+
2
e
−
(ii)
Reaction at anode in lead storage battery
(C)
2
H
2
(
g
)
+
4
O
H
−
(
a
q
)
→
4
H
2
O
(
l
)
+
4
e
−
(iii)
Electrolysis of concentrated
H
2
S
O
4
(D)
2
F
e
(
s
)
+
O
2
(
g
)
+
4
H
+
(
a
q
)
→
2
F
e
2
+
(
a
q
)
+
2
H
2
O
(
l
)
(iv)
Reaction at anode in fuel cell
Q.
Consider a galvanic cell using solid
C
u
and
F
e
metals with their corresponding solutions.
What is the
E
∘
c
e
l
l
?
Standard Potential (V)
Reduction Half-Reaction
2.87
F
2
(
g
)
+
2
e
−
→
2
F
−
(
a
q
)
1.51
M
n
O
−
4
(
a
q
)
+
8
H
+
(
a
q
)
+
5
e
−
→
M
n
2
+
(
a
q
)
+
4
H
2
O
(
l
)
1.36
C
l
2
(
a
q
)
+
3
e
−
→
2
C
l
−
(
a
q
)
1.33
C
r
2
O
2
−
7
(
a
q
)
+
14
H
+
(
a
q
)
+
6
e
−
→
2
C
r
3
+
(
a
q
)
+
7
H
2
O
(
l
)
1.23
O
2
(
g
)
+
4
H
+
(
a
q
)
+
4
e
−
→
2
H
2
O
(
l
)
1.06
B
r
2
(
l
)
+
2
e
−
→
2
B
r
−
(
a
q
)
0.96
N
O
−
3
(
a
q
)
+
4
H
+
(
a
q
)
+
3
e
−
→
N
O
(
g
)
+
H
2
O
(
l
)
0.80
A
g
+
(
a
q
)
+
e
−
→
A
g
(
s
)
$
0.77
F
e
3
+
(
a
q
)
+
e
−
→
F
e
2
+
(
a
q
)
0.68
O
2
(
g
)
+
2
H
+
(
a
q
)
+
2
e
−
→
H
2
O
2
(
a
q
)
0.59
M
n
O
−
4
(
a
q
)
+
2
H
2
O
(
l
)
+
3
e
−
→
M
n
O
2
(
s
)
+
4
O
H
−
(
a
q
)
0.54
I
2
(
s
)
+
2
e
−
→
2
I
−
(
a
q
)
0.40
O
2
(
g
)
+
2
H
2
O
(
l
)
+
4
e
−
→
4
O
H
−
(
a
q
)
0.34
C
u
2
+
(
a
q
)
+
2
e
−
→
C
u
(
s
)
0
2
H
+
(
a
q
)
+
2
e
−
→
H
2
(
g
)
−
0.28
N
i
2
+
(
a
q
)
+
2
e
−
→
N
i
(
s
)
−
0.44
F
e
2
+
(
a
q
)
+
2
e
−
→
F
e
(
s
)
−
0.76
Z
n
2
+
(
a
q
)
+
2
e
−
→
Z
n
(
s
)
−
0.83
2
H
2
O
(
l
)
+
2
e
−
→
H
2
(
g
)
+
2
O
H
−
(
a
q
)
1.66
A
l
3
+
(
a
q
)
+
3
e
−
→
A
l
(
s
)
−
2.71
N
a
+
(
a
q
)
+
e
−
→
N
a
(
s
)
−
3.05
L
i
+
(
a
q
)
+
e
−
→
L
i
(
s
)
Q.
If the equilibrium constant for the reaction :
H
+
(
a
q
.
)
+
O
H
−
(
a
q
.
)
⇌
H
2
O
(
l
)
is
10
13
at certain temperature then what is the
E
∘
for the reaction,
2
H
2
O
(
l
)
+
2
e
−
⇌
H
2
(
g
)
+
2
O
H
−
(
a
q
.
)
.
Given:
2.303
R
T
F
=
0.066
.
Q.
Consider a spontaneous electrochemical cell between
C
u
and
A
l
. Predict what would happen if excess concentrated
N
a
O
H
were added to the cell with copper ions and a precipitate forms.
Standard Potential (V)
Reduction Half-Reaction
2.87
F
2
(
g
)
+
2
e
−
→
2
F
−
(
a
q
)
1.51
M
n
O
−
4
(
a
q
)
+
8
H
+
(
a
q
)
+
5
e
−
→
M
n
2
+
(
a
q
)
+
4
H
2
O
(
l
)
1.36
C
l
2
(
a
q
)
+
3
e
−
→
2
C
l
−
(
a
q
)
1.33
C
r
2
O
2
−
7
(
a
q
)
+
14
H
+
(
a
q
)
+
6
e
−
→
2
C
r
3
+
(
a
q
)
+
7
H
2
O
(
l
)
1.23
O
2
(
g
)
+
4
H
+
(
a
q
)
+
4
e
−
→
2
H
2
O
(
l
)
1.06
B
r
2
(
l
)
+
2
e
−
→
2
B
r
−
(
a
q
)
0.96
N
O
−
3
(
a
q
)
+
4
H
+
(
a
q
)
+
3
e
−
→
N
O
(
g
)
+
H
2
O
(
l
)
0.80
A
g
+
(
a
q
)
+
e
−
→
A
g
(
s
)
$
0.77
F
e
3
+
(
a
q
)
+
e
−
→
F
e
2
+
(
a
q
)
0.68
O
2
(
g
)
+
2
H
+
(
a
q
)
+
2
e
−
→
H
2
O
2
(
a
q
)
0.59
M
n
O
−
4
(
a
q
)
+
2
H
2
O
(
l
)
+
3
e
−
→
M
n
O
2
(
s
)
+
4
O
H
−
(
a
q
)
0.54
I
2
(
s
)
+
2
e
−
→
2
I
−
(
a
q
)
0.40
O
2
(
g
)
+
2
H
2
O
(
l
)
+
4
e
−
→
4
O
H
−
(
a
q
)
0.34
C
u
2
+
(
a
q
)
+
2
e
−
→
C
u
(
s
)
0
2
H
+
(
a
q
)
+
2
e
−
→
H
2
(
g
)
−
0.28
N
i
2
+
(
a
q
)
+
2
e
−
→
N
i
(
s
)
−
0.44
F
e
2
+
(
a
q
)
+
2
e
−
→
F
e
(
s
)
−
0.76
Z
n
2
+
(
a
q
)
+
2
e
−
→
Z
n
(
s
)
−
0.83
2
H
2
O
(
l
)
+
2
e
−
→
H
2
(
g
)
+
2
O
H
−
(
a
q
)
1.66
A
l
3
+
(
a
q
)
+
3
e
−
→
A
l
(
s
)
−
2.71
N
a
+
(
a
q
)
+
e
−
→
N
a
(
s
)
−
3.05
L
i
+
(
a
q
)
+
e
−
→
L
i
(
s
)
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