Question

2.) For the following cell reaction :
Cu(s) + 2Ag" (aq) - 2Ag(s) + Cu?"(aq)
Ecell 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) 96.5 JK-1
(4) 193 JK-1​

Answer 1

Answer:

Hope it helps. Regards.

Answer 2

The entropy change will be 193 J/K

Explanation:

n=2,$T_1=300K,T_2=340K$

$E_1=0.46 V,E_2=0.50V$

$\frac{\partial E}{\partial T}=\frac{E_2-E_1}{T_2-T_1}$

$\frac{\partial E}{\partial T}=\frac{0.50-0.46}{340-300}$

$\frac{\partial E}{\partial T}=\frac{0.04}{40} V/K$

$F=96500 C/mol$

$\Delta S=nF(\frac{\partial E}{\partial T})$

Substitute the values in the formula then, we get

$\Delta S=2\times 96500\times \frac{0.04}{40}$

$\Delta S=193 JK^{-1}$

Hence, the entropy change will be 193 J/K

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