Question

2.
For the following cell reaction :
Cu(s) + 2Ag*(aq) + 2Ag(s) + Cu2+(aq)
Eºcis 0.46 V at 300 K and 0.50 V at 340 K, then
what will be entropy change?
(1) 386 JK-1
(2) 276 JK-1
(3) 96.5JK-1
(4) 193 JK-1​

Answer 1

The entropy change will be  ΔS =  - 193 . 5 JK^-1

Explanation:

Given data:

n=2

Initial temperature "T1" =300 K

Final temperature "T2" = 340 K

E1 = 0.46 V

E2 = 0.50 V

ΔS = - nF . dE / dT

ΔS = -2 x 96500 x ( 0.50 - 0.46) / 340 - 300

ΔS =  -2 x 96500 x 0.04 / 40

ΔS = -2 x 96500 x 10.3

ΔS =  -2 x 96.5

ΔS =  - 193 . 5 JK^-1

Thus the entropy change will be  ΔS =  - 193 . 5 JK^-1

Also learn more

Give reasons Water at room temperature is liquid. ​?

https://brainly.in/question/14141730

Answer 2

The change in entropy is - 193.5 JK⁻¹.

Given:

Initial temperature = T₁ =300 K

Final temperature = T₂ = 340 K

Initial emf = E₁ = 0.46 V

Final emf = E₂ = 0.50 V

Explanation:

From the given reaction, n = 2

The change in entropy is given by the formula:

ΔS = - (n × F × dE)/dT

Where

F = Faraday's constant = 96500 C mol⁻¹

n = No. of electrons involved

dT = Change in temperature

dE = Change in emf

On substituting the values, we get,

ΔS = -(2 × 96500 × (0.50 - 0.46))/(340 - 300)

ΔS =  -(2 × 96500 × 0.04)/40

ΔS = -2 × 96500 × 10.3

ΔS = -2 × 96.5

∴ ΔS = - 193.5 JK⁻¹