Based on Eo values given- state which statements are correct- 1-2 F2 - e- F- Eo - 2-87 V 1-2Cl2 - e- Cl- Eo- 1-40 V 1-2Br2 - e- Br- Eo- 1-09 V 1-2I2 - e- I- Eo - 0-62 V

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Byju's Answer

Standard XII

Chemistry

Oxoacids of Halogens

Based on Eo...

Question

Based on $E^{o}$ values given, state which statements are correct?
$\frac{1}{2} F_{2} + e^{-} ⟶ F^{-}$ $E^{o}$ = 2.87 V
$\frac{1}{2} C l_{2} + e^{-} ⟶ C l^{-}$ $E^{o}$ = 1.40 V
$\frac{1}{2} B r_{2} + e^{-} ⟶ B r^{-}$ $E^{o}$ = 1.09 V
$\frac{1}{2} I_{2} + e^{-} ⟶ I^{-}$ $E^{o}$ = 0.62 V

Oxidising power of

F_{2}

is maximum

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Oxidising power of

I_{2}

is maximum

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Oxidising power is in order

F_{2} > C l_{2} > B r_{2} > I_{2}

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Oxidising power is in order

F_{2} < C l_{2} < B r_{2} < I_{2}

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Solution

The correct options are
B Oxidising power of $F_{2}$ is maximum
D Oxidising power is in order $F_{2} > C l_{2} > B r_{2} > I_{2}$
A. The standard reduction potential of fluorine is most negative. Hence, the oxidizing power of fluorine is maximum. Thus, the option A is correct.
B. The standard reduction potential of iodine is least negative. Hence, the oxidizing power of iodine is minimum. Thus, the option B is incorrect.
C. and D. On moving down the group, the oxidizing power of halogens increases in the order $F_{2} > C l_{2} > B r_{2} > I_{2}$ . Thus, option C is correct and the option D is incorrect.

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Similar questions

C H C l_{3}

and

H_{2} O

from two separate immiscible layers.

B r_{2}

(orange) and

I_{2}

(violet) are soluble in organic layer to a greater extent. In test tube

A

(

K I + C H C l_{3} + H_{2} O

) and

B

(K B r + C H C l_{3} + H_{2} O)

C l_{2}

gas is passed then:
Given:

\frac{1}{2} C l_{2} + e^{-} ⟶ C l^{-}

E^{o}

= 1.40 V

\frac{1}{2} B r_{2} + e^{-} ⟶ B r^{-}

E^{o}

= 1.09 V

\frac{1}{2} I_{2} + e^{-} ⟶ I^{-}

E^{o}

= 0.62 V

Q. Consider the following orders –
(1)

F_{2} > C l_{2} > B r_{2} > I_{2} :

boiling point
(2)

F_{2} > C l_{2} > B r_{2} > I_{2} :

oxidizing nature
(3)

F_{2} > C l_{2} > B r_{2} > I_{2} :

EN
(4)

F_{2} > C l_{2} > B r_{2} > I_{2} :

BDE
(5)

F_{2} > C l_{2} > B r_{2} > I_{2} :

EA
(6)

F_{2} > C l_{2} > B r_{2} > I_{2} :

Reactivity
(7)

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

Acidic nature
(8)

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

Oxidizing nature
Then calculate

(x^{2} + y^{2})

when

x

is correct order and

y

is incorrect order.

Q. Based on the date given below, the correct order of reducing power is:

F e^{3 +} (a q) + e \to F e^{2 +} (a q); E^{o} = + 0.77 V

A l^{3 +} (a q) + 3 e \to A l (s); E^{o} = - 1.66 V

B r_{2} (a q) + 2 e \to 2 B^{-} (a q); E^{o} = + 1.08 V

Q. Which of the following is the cell reaction that occurs when the following half-cells are combined?

I_{2} + 2 e^{-} \to 2 I^{-} (1 M)

;

E^{o} = + 0.54

B r_{2} + 2 e^{-} \to 2 B r^{-} (1 M); E^{o} = + 1.09

Q. Given:

C o^{3 +} + e^{-} \to C o^{2 +}; E^{o} = + 1.81 V

P b^{4 +} + 2 e^{-} \to P b^{2 +}; E^{o} = + 1.67 V

C e^{4 +} + e^{-} \to C e^{3 +}; E^{o} = + 1.61 V

B i^{3 +} + 3 e^{-} \to B i; E^{o} = + 0.20 V

Oxidizing power of the species will increase in the order :

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Based on Eo values given, state which statements are correct?12F2+e−⟶F− Eo = 2.87 V12Cl2+e−⟶Cl− Eo= 1.40 V12Br2+e−⟶Br− Eo= 1.09 V12I2+e−⟶I− Eo = 0.62 V

Based on $E^{o}$ values given, state which statements are correct?
$\frac{1}{2} F_{2} + e^{-} ⟶ F^{-}$ $E^{o}$ = 2.87 V
$\frac{1}{2} C l_{2} + e^{-} ⟶ C l^{-}$ $E^{o}$ = 1.40 V
$\frac{1}{2} B r_{2} + e^{-} ⟶ B r^{-}$ $E^{o}$ = 1.09 V
$\frac{1}{2} I_{2} + e^{-} ⟶ I^{-}$ $E^{o}$ = 0.62 V