What would be the equivalent weight of the
reductant in the reaction :
[Fe(CN),13 +H,O, + 2OH- + 2[Fe(CN)]4 + 2H,0 +0,
[Given : Fe = 56, C = 12, N = 14,0 = 16, H= 1]
(1) 17
(2) 212
(3) 34
(4) 32
.
Answers
Answer:
17
Step by step explanation:
[Fe(CN)
6
]
−3
+H
2
O
2
+2OH
−
→2[Fe(CN)
6
]
4−
+2H
2
O+O
2
Assigning the oxidation state we get:
[\overset { +3 }{ Fe } (CN)_{ 6 }]^{ -3 }+H_{ 2 }\overset { -1 }{ O_{ 2 } } +2\overset { -2 }{ O } H^{ - }\rightarrow 2[\overset { +2 }{ Fe } (CN)_{ 6 }]^{ 4- }+2H_{ 2 }\overset { -2 }{ O } +\overset { 0 }{ O_{ 2 } }[
Fe
+3
(CN)
6
]
−3
+H
2
O
2
−1
+2
O
−2
H
−
→2[
Fe
+2
(CN)
6
]
4−
+2H
2
O
−2
+
O
2
0
Reduction reaction is:
Fe^{3+} \rightarrow Fe^{2+}Fe
3+
→Fe
2+
Oxidation reaction is:
2O^- \rightarrow O_22O
−
→O
2
Since, Fe gets reduced from +3 to +2 oxidation state, it is the oxidant and H_2O_2H
2
O
2
gets oxidised from -1 oxidation state to 0 it is the reductant.
Considering,
2\overset{-1}{O}^- \rightarrow \overset{0}{O_2}2
O
−1
−
→
O
2
Change in oxidation number = 2 \times 0-2 \times (-1)=22×0−2×(−1)=2
\therefore∴Equivalent mass of reductant H_2O_2=\dfrac{molar\ mass}{change\ in\ oxidation\ no.} =\dfrac {2+16 \times 2}{2}H
2
O
2
=
change in oxidation no.
molar mass
=
2
2+16×2
=\dfrac{34}{2}=17=
2
34
=17
Thus, the equivalent weight of reductant is 17.