Justify that this reaction is a redox reaction.
Calculate the oxidation number of sulphur, chromium and nitrogen in H,SO,
Cr,o and Nog. Suggest structure of these compounds. Count for the fallacy.
Answers
Explanation:
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(a) H
2
SO
5
by conventional method.
Let x be the oxidation number of S
2(+1)+x+5(−2)=0
x=+8
+8 Oxidation state of S is not possible as S cannot have oxidation number more than 6. The fallacy is overcomed if we calculate the oxidation number from its structure HO−S(O
2
)−O−O−H.
−1+X+2(−2)+2(−1)+1=0
x=+6
(b) Dichromate ion
Let x be the oxidation number of Cr in dichromate ion
2x+7(−2)=−2
x=+6
Hence the oxidation number of Cr in dichromate ion is +6. This is correct and there is no fallacy.
(c) Nitrate ion, by conventional method
Let x be the oxidation number of N in nitrate ion.
x+3(−2)=−1
From the structure O
−
−N
+
(O)−O
−
x+1(−1)+1(−2)+1(−2)=0
x=+5
Thus there is no fallacy.