Consider the following cell:
$P t, H_{2} (g) | 2 M C H_{3} C O O N H_{4} (a q) | | 2 M N a C l (a q) | H_{2} (g), P t$
20 atm 0.2 atm
Given, $p K_{a} o f C H_{3} C O O H = 4.74 a n d p K_{b} o f N H_{4} O H i s 4.74$ .
If $E$ is e.m.f. of the cell in volt, then the value of 1000E is [Take $\frac{2.303 R T}{F} = 0.059$ ](e.g., if the value is 56, then write answer as 6)

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Solution

Since, the pKa of acetic acid is equal to the pKb of ammonium hydroxide, the solution of ammonium acetate (a salt of weak acid with weak base) will be neutral.
NaCl is a salt of strong acid with strong base. Hence, its solution will be neutral.
Thus, in both the cases, the hydrogen ion concentration will be $10^{- 7} M$ .
The value of the reaction quotient will be $Q = \frac{(10^{- 7})^{2}}{20} \times \frac{0.2}{(10^{- 7})^{2}} = 0.01$
The expression for the cell potential is
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.059}{n} l o g Q$
Substitute values in the above expression.
$E_{c e l l} = 0.0 + \frac{0.059}{2} l o g 0.01 = 0.059$
Hence, $1000 E = 1000 \times 0.059 = 59$ .

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P t, H_{2} (g) | 2 M C H_{3} C O O N H_{4} | | 2 M N a C l | H_{2} (g), P t

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p K_{b} (N H_{4} O H) = 4.74

\frac{2.303 R T}{F} = 0.059)

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