24Ans1 4n+2 where n=0,1,2,.................
Ans2 +2
A compound AX2 reacts with another compound BY in a redox reaction.X is an element in AX2 with oxidation number +3.Y is an element in BY with oxidation number -2.In one of the products CX , X has oxidation number +1.The other product is DY.A,B,C,D are the remaining atoms in these compounds,none of which undergo redox reaction.
AX2 +BY--------->CX+DY
Q1 What are posssible oxidation numbers of Y in DY?????
Q2 If the coefficients of AX2 and BY in the balanced reaction are both equal to one,what is oxidation number of Y in DY??
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UP 0 DOWN 0 0 3
3 Answers
1357take half reactions as
(AX2+4e-→2CX)×m (I)
let the oxidation state of Y be n
(BY→DY+(n+2)e-)×k (II)
adding (I) & (II) we get
4m=k(n+2)
n=-2+4m/k
for different values of m & k
we get n=-1,0,1,2,3,4,5,6,7,8 (8 is the max oxidation no. an atom can shown)
now when k=m
we get n=2
1357In Q1 X is reduced from +3 to +1. So Y has to be oxidized
So theoretically it can take any values greater than -2
but as it forms a compound with D the valency of Y should be an integer in DY
so the integers greater than -2 are the answers
but as oxidation no. cannot be greater than 8
so the answer is
n-2 (n=1,2,3.....10)
and for second I have already given the answer 2