The cell potential for the given cell at 298 K

$P t | H_{2} (g, 1 b a r) | H^{+} (a q) | | C u^{2 +} (a q) | C u (s) i s 0.31 V .$ The pH of the acidic solution is found to be 3, whereas the concentration of Cu²⁺ is 10^-x M. The value of x is________.

(Given : $E_{C u^{2 +} / C u}^{Θ} = 0.34 V a n d \frac{2.303 R T}{F} = 0.06 V)$

2 Views|Posted 4 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

4 months ago

H₂ (g) + Cu²⁺ (aq) $\overset{}{\to}$ 2H⁺ (aq) + Cu (s)

$∴ E_{c e l l} = E_{c e l l}^{0} - \frac{2.303 R T}{n f} l o g Q$

0.31 = 0.34 - $\frac{0.06}{2} l o g \frac{[H^{+}]}{C u^{2 +}}$

$[C u^{2 +}] = 1 0^{- 7} (? [H^{+}] = 1 0^{- 3})$

x = 7

Upvote

Similar Questions for you

We are given with following cell reaction:

2H⁺ + 2e^–  $\overset{}{\to}$ H₂

_{$P_{H_{2}}$ =}_{2 atm}

[H⁺] = 1 M

$(\frac{2.303 R T}{F} = 0.06)$

If E_cell for reaction is given by –x × 10^–3 V, find out x.

Alok Kumar Singh

E_cell = 0 – log2

= – 0.03 (0.3)

= – 0.009

= – 9 * 10^–3 V

x = 9

Vishal Baghel

H₂ (g) + Cu²⁺ (aq) $\overset{}{\to}$ 2H⁺ (aq) + Cu (s)

$∴ E_{c e l l} = E_{c e l l}^{0} - \frac{2.303 R T}{n f} l o g Q$

0.31 = 0.34 - $\frac{0.06}{2} l o g \frac{[H^{+}]}{C u^{2 +}}$

$[C u^{2 +}] = 1 0^{- 7} (? [H^{+}] = 1 0^{- 3})$

x = 7

For the cell Cu(s) | Cu²⁺ (aq)(0.1M) || Ag⁺ (aq)(0.01M) | Ag(s)
The cell potential E₁ = 0.3095V
For the cell Cu(s) | Cu²⁺ (aq)(0.01M) || Ag⁺ (aq)(0.001M) | Ag(s)
The cell potential = _____ ×10⁻² V. (Round off to the Nearest Integer).
[Use : (2.303RT/F) = 0.059 ]

Alok Kumar Singh

E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
E? = E? _cell - (0.059/2)log [0.1/ (0.01)²] = 0.3095
E? _cell = 0.3095 + 0.0885 V
E? _cell = 0.398

Again, E? = E? _cell - (0.059/2)log [10? ²/ (10? ³)²]
E? = 0.398 - (0.059/2) x 4 V

E? = 0.28 V
E? = 28 x 10? ² V

Find the emf of the cell in which the following reaction takes place at 298 K
Ni(s) + 2Ag⁺(0.001M) → Ni²⁺(0.001M) + 2Ag(s)
(Given that E°cell = 1.05 V, 2.303RT/F = 0.059 at 298 K)

Vishal Baghel

E_cell = 1.05 - (0.059 / 2) log ( [Ni²? ] / [Ag? ]²)
= 1.05 - (0.059 / 2) log ( [10? ³] / [10? ³]²)
= 1.05 - (0.059 × 3) / 2 = 1.05 - 0.0885 = 0.9615 volt
There is a misprint in the question. The E? _cell is incorrectly given as 10.5 V. This should have been 1.05 volt.

Raj Pandey