Question

i)

Consider the following cell reaction at 298 K: [3]
2Ag⁺ + Cd → 2Ag + Cd²⁺
The standard reduction potentials (E°) for Ag⁺/Ag and Cd^2-/Cd are 0.80 V and -0.40 V respectively:
(1) Write the cell representation.
(2) What will be the emf of the cell if the concentration of Cd²⁺ is 0.1M and that of Ag⁺ is 0.2 M?
(3) Will the cell work spontaneously for the condition given in (2) above?

(ii)

What is a buffer solution? How is it prepared? Explain the buffer action of a basic buffer with a suitable example.

Answer 1

(i)

(1)

Cd | Cd²⁺ (C₁M) || (C₂M)Ag⁺ | Ag

e.g.,

Cd | Cd²⁺ (0.1M) || (0.2)Ag⁺ | Ag

Cell reaction is :

\[\ce{Cd->Cd^2+ +2e^-}\]

\[\ce{2Ag^+ +2e^- -> 2Ag}\]

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\[\ce{Cd + 2Ag^+ -> Cd^2+ + 2Ag}\]
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 (2)

n = 2

T = 298K

Nernst equation for the cell reaction is

`E_(cell) = E_"cell"^circ + 0.0591/n log [Ag^+]^2/[Cd^(2+)]`

`E_"cell"^circ = 0.80 + 0.40 = 1.20`V

[Ag⁺] = 0.2 M

[Cd²⁺] = 0.1 M

∴ `E_(cell) = 1.20 + 0.0591/2 log  (0.2)^2/0.1`

= `1.20 + 0.0591/2 log  0.4`

= `1.20 + 0.0591/2 xx 0.6021`

= 1.200 + 0.018

= 1.218 V

\[\ce{\underset{\text{(From HCl)}}{H^+} + \underset{\text{(From buffer)}}{OH^-} -> {H_2O}}\]

(3)

Since E_cell is +ve, the cell will work spontaneously.

(ii) 

Buffer solution: It is a solution whose pH remains unchanged on the addition of a small amount of acid or base or on dilution.
Preparation of buffer solution: A buffer solution can be prepared by mixing weak acid and its salt with a strong base or a weak base and its salt with strong acid.
The action of a basic buffer
Let us consider a basic buffer obtained by mixing solutions containing 1 mole of NH₄OH and 1 mole of NH₄Cl. Its pH value is nearly 9.25.
In the solution, there will be a large concentration of `"NH"_4^+` and Cl^– ions and unionised NH₄OH molecules.

Suppose a few drops of HCl are added to the buffer. Here H⁺ ions from HCl will combine with OH^– ions from NH₄OH to form unionised water molecules.

\[\ce{\underset{\text{Bronsted acid}}{NH4OH} ⇌ \underset{\text{Bronsted base}}{NH4+ + OH-}}\]

According to Le-Chatelier’s principle, due to the removal of OH ions, the equilibrium (1) will shift towards r.h.s., giving more OH^– ions and original concentration of OH^– ions is restored. Therefore, the pH of the solution remains unchanged.

Suppose a few drops of NaOH are added to the buffer. Here OH^– ions from NaOH combine with NH⁴⁺ ions from NH₄Cl to give weakly ionised NH₄OH molecules.

\[\ce{\underset{\text{(From buffer)}}{NH_4^+} + {OH^-} -> {NH_4OH}}\]

Suppose the buffer solution is diluted with water, it will not change the pH of the solution because the concentration of H⁺ and OH^– ions remains unchanged.