For the cell, $P t | C l_{2} (g, 0.4 b a r) | C l^{-} (a q, 0.1 M) | | C l^{-} (a q, 0.01 M) | C l_{2} (g, 0.2 b a r) | P t$ . The cell potential at $298 K$ is :

0.051 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 0.051 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.102 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0255 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.051 V$
Anode: $2 C l^{-} (a q) \to C l_{2} (g) + 2 e^{-}$
Cathode: $C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q)$
Cell reaction: $2 C l^{-} (a q) + C l_{2} (g) \to 2 C l^{-} (a q) + C l_{2} (g)$
$E_{c e l l} = - \frac{0.06}{2} l o g \frac{[C l^{-}]_{R H S}^{2} . (P_{C l_{2}})_{L H S}}{[C l^{-}]_{L H S}^{2} . (P_{C l_{2}})_{R H S}}$ $(∵ E_{c e l l}^{\circ} = 0)$
$= - \frac{0.06}{2} l o g \frac{(0.01)^{2}}{(0.1)^{2}} \times (\frac{0.4}{0.2}) \Rightarrow 0.051 V$

Hence option (A) is correct.

Suggest Corrections

Similar questions

25^{o} C

P t | {B r}_{2} (g) | {B r}^{-} (a q) | | {C l}^{-} (a q) | {C l}_{2} (g) | P t

25^{\circ} C

P t (s) | B r^{-} (a q) | B r_{2} (g) | | C l_{2} (g) | C l^{-} (a q) | P t (s)

P t | H_{2} (g, 0.1 b a r) | H^{+} (a q, 10^{- 3} M) | | M n O_{4}^{-} (a q, 0.1 M)

M n^{2 +} (a q, 0.01 M), H^{+} (a q, 0.01 M) | P t

E_{M n O_{4}^{-} | M n^{2 +}}^{\circ} = 1.51 V

P t | O_{2} (g, 1 a t m) | O H^{-} (1 M) | | C l^{-} (1 M) | C l_{2} (g, 1 a t m) | P t

P t | C l_{2} (g, 0.5 bar) | C l^{-} (aq, 0.5 M) | | C l^{-} (aq, 0.05 M) | | C l_{2} (g, 0.05 bar) | P t

298

Join BYJU'S Learning Program