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Byju's Answer
Standard XII
Chemistry
Oxidising Reagent and Reducing Reagent
Consider the ...
Question
Consider the following statements for the reaction:
K
I
O
3
+
5
K
I
+
6
H
C
l
→
3
I
2
+
6
K
C
l
+
3
H
2
O
1.
I
−
is reduced to
I
2
2.
I
O
3
−
is oxidized to
I
2
3.
I
O
3
−
is reduced to
I
2
4.
Oxidation number of
I
increases from
−
1
(in
K
I
) to zero (in
I
2
)
The correct statements are:
A
3
and
4
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B
1
and
4
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C
1
,
3
and
4
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D
1
,
2
and
4
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Solution
The correct option is
A
3
and
4
K
I
+
5
O
3
+
5
K
I
−
1
+
6
H
C
l
→
3
I
2
0
+
6
K
C
l
+
3
H
2
O
I
−
is oxidized to
I
2
.
I
O
−
3
is reduced to
I
2
.
Oxidation number of
I
increases from
−
1
(in
K
I
) to
z
e
r
o
(in
I
2
).
Hence, option
A
is correct.
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1
Similar questions
Q.
Select the oxidant in the given reactions.
K
I
O
3
+
5
K
I
+
6
H
C
l
→
3
I
2
+
6
K
C
l
+
3
H
2
O
Q.
Oxidation number of iodine in
I
O
−
3
,
I
O
−
4
,
K
I
and
I
2
respectively is _____
Q.
In the following Reaction, which substance is oxidised and reduced?
I
2
+
N
a
O
H
→
I
O
3
+
N
a
I
+
3
H
2
O
Q.
Assertion :
K
I
O
3
reacts with
K
I
to liberate iodine which is titrated with standard hypo solution. The reactions are:
(i)
I
O
−
3
→
I
2
(Valency factor = 5/3)
(ii)
I
2
+
S
2
O
2
3
→
S
4
O
2
−
6
+
I
−
(Valence factor
=
2)
Milliequivalent
of hypo
=
meq. of
I
2
=
meq. of
I
O
−
3
.
Milliequivalent
of hypo =
2
×
m
e
q
.
of
I
O
−
3
=
meq. of
I
−
.
Reason: Valency factor of
I
2
in both the equation are different, therefore we cannot compare milliequivalents in sequence.
Q.
Consider the disproportionation of Iodine to iodide and iodate ions. Using Oxidation method calculate the ratio of iodate and iodide ions formed in alkaline medium :
I
2
→
I
O
−
3
+
I
−
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Standard XII Chemistry
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