The thing you MUST do FIRST is look for any H's, O's, or F's in the equation
1)any element just by itself not in a compound, their oxidation number is 0
ex: H2's oxidation number is 0
ex: Ag: oxidation number is 0 if its just something like Ag + BLA = LALA
2) the oxidation number of H is always +1, unless its just by itself (see #1)
3) the oxidation number of O is always -2, unless its just by itself (see #1)
4) the oxidation number of F is always -1, unless its just by itself (see#1)
ok so after you have written those oxidation numbers in rules 1-4 over each H, F, or O atom in the compound, you can look at the elements that we havent talked about yet
for example::::
N2O4
the oxidation number of O is -2.
since there are 4 O's, the charge is -8. now remember that N2O4 has to be neutral so the N2 must have a charge of +8
+8 divided by 2 = +4
N has an oxidation number of +4.
more rules:
5) the sum of oxidation numbers in a compound add up to 0 (when multiplied by the subscripts!!!) (see above example)
6) the sum of oxidation numbers in a polyatomic ion is the charge (for example, PO4 has a charge of (-3) so
oxidation # of O = -2. (there are 4 O's = -8 charge on that side ) P must have an oxidation number of 5. (-8+5= -3), and -3 is the total charge of the polyatomic ion
Mole ratio of H₂ : O₂ is 2 : 1
∴ if mole of H₂ = 1.67 mol
then mole of O₂ =
⇒ that mole of oxygen needed to react with
H₂ = 0.835 molThe answer is A
Answer:
Oxidizing agent - CrO4^2-
Reducing agent- N2O
Explanation:
Let us look at the equation closely;
CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]
The reduction half equation is;
CrO4^2- (aq) + 3e -------->Cr^3+ (aq)
Oxidation half equation is;
3N2O(g) ------>3 NO(g) +3 e
Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.
The answer would most likely be true.
