Advertisement Remove all ads

Justify Giving Reactions that Among Halogens, Fluorine is the Best Oxidant and Among Hydrohalic Compounds, Hydroiodic Acid is the Best Reductant. - Chemistry

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Advertisement Remove all ads

Solution 1

F2 can oxidize Cl– to Cl2, Br to Br2, and I to I2 as:

`F_(2(aq)) + 2Cl_(s)^(-) -> 2F_(aq)^-  + Cl_(g)`

`F_(2(aq)) + 2Br_(aq)^- -> 2F_(aq)^(-) + Br_(2(/))`

`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.

`2HI + H_2SO_4 -> I_2 + SO_2 + 2H_2O`

`2HBr + H_2SO_4 -> Br_2 + SO_2 + 2H_2O`

Again, I can reduce Cu2+ to Cu+, but Br cannot.

`4I_(aq)^-  + 2Cu_(aq)^(2+) -> 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.

Solution 2

Halogens have a strong tendency to accept electrons. Therefore, they are strong oxidising agents. Their relative oxidising power is, however, measured in terms of their electrode potentials. Since the electrode potentials of halogens decrease in the order: F2(+2.87V) > Cl2 (+1.36V) > Br2 (+1.09V) > I2 (+0.54V), therefore, their oxidising power decreases in the same order

This is evident from the observation that F2 oxidises Cl to Cl2, Brto Br2, I – to I2 ; Cl2 oxidises Brto Br2 and F to I2 but not F to F2. Br2, however, oxidises F to I2 but not F–  to F2 , and Cl–   to Cl2

F2(g) + 2Cr(aq) ———–> 2F(aq) + Cl2(g); F2(g) + 2Br(aq) ———-> 2F(aq) + Br2 (Z)

F2(g) + 2I(aq) ———-> 2F(aq) + I2(s); Cl2 (g) + 2Br(aq) ————> 2Cl(aq) + Br2 (Z)

Cl2(g) + 2I(aq) ———–> 2Cl–  (aq) + I2(s) and Br2 (Z) + 2F ———> 2Br(aq) + I2(s)

Thus, F2 is the best oxidant.
Conversely, halide ions have a tendency to lose electrons and hence can act as reducing agents. Since the electrode potentials of halide ions decreases in the order: I(-0.54 V) > Br (-1.09 V) > Cl(-1.36 V) > I(-2.87 V), therefore, the reducing power of the halide ions or their corresponding hydrohalic acids decreases in the same order: HI > HBr > HCl > HF. Thus, hydroiodic acid is the best reductant. This is supported by the following reactions. For example, HI and HBr reduce H2S0to S0while HCl and HF do not.

2HBr + H2S04 —–> Br2+ S02 + 2H2O; 2HI + H2S04 ——> I2 + S02 + 2H2O
Further F reduces Cu2+ to Cu+ but Br does not.
2Cu2+(aq) + 4I(aq) >Cu2I2(s) + I2(aq); Cu2+(aq) + 2Br> No reaction.
Thus, HI is a stronger reductant than HBr.
Further among HCl and HF, HCl is a stronger reducing agent than HF because HCl reduces MnOto Mn2+ but HF does not.
MnO(s) + 4HCl(aq) ——-> MnCl2(aq) + Cl2(aq) + 2H2O
MnO(s) + 4HF(l) ———–> No reaction.
Thus, the reducing character of hydrohalic acids decreases in the order: HI > HBr > HCl > HF

Concept: Oxidation Number - Redox Reactions as the Basis for Titrations
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 8 Redox Reactions
Q 15 | Page 273
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×