Answer to Question #161119 in Physical Chemistry for su

Question

The e.m.f. of the following cell reaction is 0.6753 V at 25 ℃ and 0.6915 V at 0 ℃. Cd(s) | CdCl2.5H2O(saturated) || AgCl(s) | Ag(s) (i) Give the balance for the cell reaction (ii) Calculate the free energy G, enthalpy H and entropy S, change of the cell reaction at 25 ℃. 

Answer 1

1. Balanced equation for the cell reaction is:

Cd(s) + 2AgCl(s) + 5H2O(l) --------> CdCl2.5H2O(s) + 2Ag(s)

2. Free energy at 25°C,∆G= ?

number of moles of electrons transferred= 2

Faraday's constant, F= 96500C

Emf, E= 0.6753V

∆G= -nFE

∆G= - 2 x 96500 x 0.6573

∆G= -126858.9J

∆G= -127Kj

To find ∆H and ∆S, let's find ∆G at 0°C and set a simultaneous equation

∆G= -nFE

Where E = 0.6915

∆G= -2 x 96500 x 0.6915= -133Kj

Now from the equation

∆G= ∆H - T∆S

Let ∆H be x and ∆S be y. We can now set two equations at 25°C and 0°C

-127= x - 298y.......(I)

-133= x - 273y........(II)

Subtract (I) and (II) to eliminate x

6= -25y

y= -6/25 = -0.24Kj

Substitute in (I)

-127= x - 298(-0.24)

-127= x + 71.52

x= -127-71.52

x= -198.52Kj

Therefore

x= ∆H= -198.52Kj

y= ∆S= -0.24Kj= -240J