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Byju's Answer
Standard XII
Chemistry
Hydrogen Peroxide
Zn acts as sa...
Question
Zn acts as sacrificial or Cathodic protect ion to prevent rusting of iron because:
A
E
⊖
O
P
of Zn <
E
⊖
O
P
of Fe
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B
E
⊖
O
P
of Zn >
E
⊖
O
P
of Fe
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C
E
⊖
O
P
of Zn =
E
⊖
O
P
of Fe
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D
Zn is cheaper than iron
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Solution
The correct option is
B
E
⊖
O
P
of Zn >
E
⊖
O
P
of Fe
E
o
o
p
>
E
o
p
(
Z
n
)
(
F
e
)
that means oxidation of
Z
n
→
Z
n
2
+
will occur instead of
F
e
→
F
e
2
+
/
F
e
3
+
, thus collosion is protected.
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Similar questions
Q.
Find the
E
0
c
e
l
l
for the following cell reaction:
F
e
=
2
+
Z
n
→
Z
n
+
2
+
F
e
Given,
E
0
Z
n
/
Z
n
+
2
=
0.76
V
,
E
0
F
e
/
F
e
+
2
=
+
0.41
V
Q.
The standard reduction potential
E
o
for half reaction are:
Z
n
→
Z
n
2
+
+
2
e
−
;
E
o
=
+
0.76
V
F
e
→
F
e
2
+
+
2
e
−
;
E
o
=
+
0.41
V
The EMF of the cell reaction is:
F
e
2
+
+
Z
n
→
Z
n
2
+
+
F
e
Q.
The standard reduction potential
E
o
for the half reactions are as
Z
n
⟶
Z
n
2
+
+
2
e
−
;
E
o
=
+
0.76
V
F
e
⟶
F
e
2
+
+
2
e
−
;
E
o
=
+
0.41
V
The emf for cell reaction,
F
e
2
+
+
Z
n
⟶
Z
n
2
+
+
F
e
, is
Q.
The standard oxidation potentials,
E
∘
, for the half reactions are as,
Z
n
→
Z
n
2
+
+
2
e
−
;
E
∘
=
+
0.76
v
o
l
t
F
e
→
F
e
2
+
+
2
e
−
;
E
∘
=
+
0.41
v
o
l
t
The emf of the cell,
F
e
2
+
+
Z
n
→
Z
n
2
+
+
F
e
is:
Q.
Given the standard half-cell potentials
(
E
o
)
of the following as
Z
n
=
Z
n
2
+
+
2
e
−
;
E
o
=
+
0.76
V
F
e
=
F
e
2
+
+
2
e
−
;
E
o
=
+
0.41
V
Then the standard e.m.f. of the cell with the reaction
F
e
2
+
+
Z
n
→
Z
n
2
+
+
F
e
is?
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