The EMF of the following concentration cell is 0.72 V. $P t | H_{2} (1 a t m) | K O H (0.1 M) | | H C l (0.48 M) | H_{2} (1 a t m), P t$ . Assuming that KOH is 85% ionized, if the ionic product of water is $2.55 \times 10^{- x}$ , the value of x is.............

Open in App

Solution

$\begin{matrix} K O H 0.1 M \end{matrix} ⇌ \begin{matrix} K^{\oplus} - \end{matrix} + \begin{matrix} ⊖ O H - \end{matrix}$
KOH is 85% ionizable.
$\Rightarrow [⊖ O H] = 0.1 \times 0.85 = 0.085 M$
$[H^{\oplus}] = \frac{K w}{[⊖ O H]} = c_{1} (s a y)$
Anode : $\frac{1}{2} H_{2} \to H^{\oplus} (c_{1}) + e^{-}$
Cathode : $H^{\oplus} + e^{-} \to \frac{1}{2} H_{2}$
$E_{c e l l} = 0.72 V$
$\Rightarrow 0.72 = - \frac{0.059}{1} l o g \frac{c_{1}}{0.48}$
$\Rightarrow \frac{- 0.72}{0.059} = l o g \frac{K w}{0.085 \times 0.48}$
$\Rightarrow K w = 2.55 \times 10^{- 14}$
so value of x is 14.

Suggest Corrections

Similar questions

P t, H_{2} (1 a t m) | H C l (0.1 M) | | {C H}_{3} C O O H (0.1 M) | H_{2} (1 a t m), P t

P t | H_{2} (1 a t m) | H C l (0.01 M) | | H C l (p H = X) | H_{2} (0.01 a t m) | P t

$E^{0} Z n^{2 +} / Z n = - 0.76 V$ The EMF of the cell

$Z n / Z n^{2 +} (1 M) | | H C l (p H = 2) | H_{2 (1 a t m)}, P t$ is ::

E_{c e l l} = 0.2364 V

P t, (H_{2}) (1 a t m) | H^{\oplus} (x M) | | H^{\oplus} (1 M) (1 a t m) | P t, (H_{2})

M / 32

P t | H_{2} (1 a t m) | H^{+} (1 M) | | M / 32 C_{6} H_{6} N H_{3} C l | H_{2} (1 a t m) | P t; E_{c e l l} = - 0.188 V

h = 2.12 \times 10^{- 2}, K_{h} = 1.43 \times 10^{- 4} M

Join BYJU'S Learning Program