Advertisements
Advertisements
Question
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Solution
F2 can oxidize Cl– to Cl2, Br– to Br2, and I– to I2 as:
\[\ce{F_{2(aq)} + 2Cl-_{(s)} -> 2F-_{(aq)} + Cl_{(g)}}\]
\[\ce{F_{2(aq)} + 2Br-_{(aq)} -> 2F-_{(aq)} + Br_{2(l)}}\]
\[\ce{F_{2(aq)} + 2I-_{(aq)} -> 2F-_{(aq)} + I_{2(s)}}\]
On the other hand, Cl2, Br2, and I2 cannot oxidize F– to F2. The oxidizing power of halogens increases in the order of I2 < Br2 < Cl2 < F2. Hence, fluorine is the best oxidant among halogens.
HI and HBr can reduce H2SO4 to SO2, but HCl and HF cannot. Therefore, HI and HBr are stronger reductants than HCl and HF.
\[\ce{2HI + H_2SO_4 -> I_2 + SO_2 + 2H_2O}\]
\[\ce{2HBr + H_2SO_4 -> Br_2 + SO_2 + 2H_2O}\]
Again, I– can reduce Cu2+ to Cu+, but Br– cannot.
\[\ce{4I-_{(aq)} + 2Cu^{2+}_{(aq)} -> Cu_2I_{2(s)} + I_{2(aq)}}\]
Hence, hydroiodic acid is the best reductant among hydrohalic compounds.
Thus, the reducing power of hydrohalic acids increases in the order of HF < HCl < HBr < HI.
APPEARS IN
RELATED QUESTIONS
Given below are two statements:
Statements I: In the titration between strong acid and weak base methyl orange is suitable as an indicator.
Statement II: For titration of acetic acid with NaOH, phenolphthalein is not a suitable indicator.
In light of the above statements, choose the most appropriate answer from the options given below:
A 0.5 g sample of an iron-containing mineral mainly in the form of CuFeS2 was reduced suitably to convert all the ferric iron into ferrous form and was obtained as a solution. In the absence of any interfering matter, the solution required 42 ml of 0.01 M K2Cr2O7 solution for titration. The percentage of CuFeS2 in the mineral is ______%.
(Cu = 63.5, Fe = 55.8, S = 32, O = 16)
