The pH in LHE of the following cell is, if $E_{c e l l} = 0.2364 V$ .

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

Open in App

Solution

Anode : $H_{2} (1 a t m) \to 2 H^{\oplus} (x M) + 2 e^{-}$

Cathode : $2 H^{\oplus} (1 M) + 2 e^{-} \to H_{2}, (1 a t m)$

$0.2364 = 0 - \frac{0.059}{2} l o g {(\frac{x}{1})}^{2}$

$\Rightarrow \frac{0.2364}{0.059} = - l o g x =$ $p H$

$\Rightarrow$ pH = 4

Suggest Corrections

Similar questions

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

p t, H_{2} (1 a t m) | H^{\oplus} (c) | | K C l (s a t u r a t e d) | H g_{2} C l_{2} | H g (s)

0.36

0.242

H^{\oplus}

P t (H_{2}) | H^{+} (p H = 2) | H^{+} (p H = 3) | P t (H_{2})

P t | H_{2} (x a t m) | H^{\oplus} (p H = y) | | Z n^{2 +} (z M) | Z n

E_{c e l l}^{⊖} = E_{c e l l}

I_{2} (S) | I (0.1 M)

H_{(a q)}^{+} | H_{2} (1 a t m) | P t

E_{I_{2} / I}^{o} = 0.535 V

H^{+} / H_{2}

Join BYJU'S Learning Program