Tardigrade
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Tardigrade
Question
Chemistry
In the reaction 2 Na2S2O3 + I2 arrow Na2 S4 O6 + NaI the oxidation number of sulphur is
Q. In the reaction
2
N
a
2
S
2
O
3
+
I
2
→
N
a
2
S
4
O
6
+
N
a
I
the oxidation number of sulphur is
1541
222
Redox Reactions
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A
decreased
18%
B
increased
51%
C
unchanged
23%
D
none of these
7%
Solution:
N
a
2
S
2
O
3
=
+
2
+
2
x
−
6
=
0
or
2
x
=
4
or
x
=
+
2
.
N
a
2
S
4
O
6
=
+
2
+
4
x
−
12
=
0
or
4
x
=
10
or
x
=
4
10
=
2.5
Here, the O.N. of S is increased.