Cds - CdCl20-10 M - AgCls - Ags The EMF of the above cell is 0-6315 V at 0-C and 0-6753 V at 25-C- The - H of reaction in kJ at 25-C is-

The correct option is D 167.6Note : Cd(s) | CdCl_2(0.1 #160;M) | AgCl(s) | Ag(s) is equivalent to : Cd(s) | Cd^2+(0.1 M) || Cl^⊖(0.2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Nernst Equation

Cds | CdCl20....

Question

$C d (s) | C d C l_{2} (0.10 M) | A g C l (s) | A g (s)$

The EMF of the above cell is 0.6315 V at $0^{\circ} C$ and 0.6753 V at $25^{\circ} C$ . The $Δ H$ of reaction in kJ at $25^{\circ} C$ is:

- 176

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

- 234.7

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

123.5

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

- 167.6

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

Open in App

Solution

The correct option is D - 167.6
Note : $C d (s) | C d C l_{2} (0.1 M) | A g C l (s) | A g (s)$ is equivalent to : $C d (s) | C d^{2 +} (0.1 M) | | C l^{⊖} (0.2 M) | A g C l (s) | A g (s)$
Use : $Δ H = n F G [T . (\frac{d E_{c e l l}}{d T}) - 0.6753]$
$Δ H_{25^{\circ} C} = 2 \times 96500 [298 \times (\frac{0.6753 - 0.6915}{25 - 0}) - 0.6753]$
$= - 167.67 k J m o l^{- 1}$

Suggest Corrections

Similar questions

Q. The EMF of the cell,

C d (s) | C d C l_{2} (a q, 0.1 M) | | A g C l (s) | A g (s)

in which the cell reaction is

C d (s) + 2 A g C l (s) \to 2 A g (s) + C d^{2 +} (a q) + 2 C l^{-} (a q)

0.6915 V

at 0°C and

0.6753 V

at 25°C. The

Δ S

and

Δ H

of the reaction at 25°C is:
Take,

F = 96485 C m o l^{- 1}

Q. Emf of the cell

P t . H_{2} (1 a t m) | H^{+} (a q .) | | A g C l | A g

0.27 V

and

0.26 V

25^{\circ} C

and

35^{\circ} C

. Heat of reaction occuring inside the cell at

25^{\circ} C

is:

The standard potential of the following cells is 0.23 V at 15

^{o}

C and 0.21 V at 35

^{o}

P t | H_{2} (g) | H C l (a q) | A g C l (s) | A g (s)

The solubility of

A g C l

in water 25

^{o}

[Note : The standard reduction potential of the

A g^{+} (a q) / A g (s)

couple is 0.80 V at 25

^{o}

Q. The EMF of the cell :

P t,

H_{2} (1 a t m) | H^{\oplus} (a q) ∥ A g C l | A g

is 0.27 and 0.26 V at

25^{\circ} C

and

35^{\circ} C

, respectively. The posituve value of heat of the reaction occurring inside the cell at

25^{\circ} C

(

i n k J K^{- 1}

) is: ( write your answer to nearest integer)

Q. The standard emf of the cell

C d (s) / C d C l_{2} (a q) (0.1 M) | | A g C l (s) | A g (s)

in which the cell-reaction is

C d (s) + 2 A g C l (s) \to 2 A g (s) + C d^{2 +} (a q) + 2 C l^{-} (a q)

0.6915

V at

273

K and

0.6573

V at

298

K. Calculate enthalpy change of the reaction at

298

K is?

Join BYJU'S Learning Program

Cd(s)|CdCl2(0.10M)|AgCl(s)|Ag(s) The EMF of the above cell is 0.6315 V at 0∘C and 0.6753 V at 25∘C. The ΔH of reaction in kJ at 25∘C is:

The correct option is D - 167.6Note : Cd(s)|CdCl2(0.1M)|AgCl(s)|Ag(s) is equivalent to : Cd(s)|Cd2+(0.1M)||Cl⊖(0.2M)|AgCl(s)|Ag(s)Use : ΔH=nFG[T.(dEcelldT)−0.6753]ΔH25∘C=2×96500[298×(0.6753−0.691525−0)−0.6753]=−167.67kJmol−1

$C d (s) | C d C l_{2} (0.10 M) | A g C l (s) | A g (s)$

The EMF of the above cell is 0.6315 V at $0^{\circ} C$ and 0.6753 V at $25^{\circ} C$ . The $Δ H$ of reaction in kJ at $25^{\circ} C$ is: