The E os for the following cell at 298 K and 308 K are 0.058V and 0.0614 V respectively. What is the cell reaction? Calculate G o , S o and H o at 298 K. The cell is: Ag (s) | AgCl (s) | KCl (1M) |Hg2 Cl2 (s) | Hg ( l )
Solution
The reactions at the cathode (right electrode) and anode (left electrode) are respectively.
Cathode: Hg2Cl2 (s) + 2e 2Hg (l) + 2 Cl - (aq).
Anode: 2Ag(s) + 2 Cl -(aq) 2 AgCl + 2e
Adding these two, we get the cell reaction, 2Ag(s) + Hg2Cl2 2 AgCl + Hg (l)
G o = - n Fo = -2 x 96500 x 0.058 = -11.194 kJ for the reaction given above.
To obtain S o , we need dE / dT at 298 K.
The correct way of obtaining dE / dT at a temperature T 0 is to obtain the values of E at several values of T around T 0 and plot E vs T and obtain the tangent to the curve at T = T 0 . It is not practical to have thermostats at every 1 degree (Celsius) or 0.1 degree interval. Since the emf generally does not vary by more than 0.001 V for a change in temperature of one degree, we may take dE /dT to be E / T = (0.0614 – 0.058) / 10 = 3.4 x 10 - 4 V / K at all temperatures in the range from 298 K to 308 K.
S o = n F d o / dT = 2 x 96500 x 3.4 x 10 – 4
= 57.9 VC / K = 57.9 J / K
Finally H o = G o + T S o
= -11,194 + 278 x 57.9
= 6060 J
The values of G o , H o and S o are for the cell reaction as written above wherein two moles of electrons are transferred (n = 2) and therefore, the phrase per mole is excluded in the units.
G o , 
2Hg (l) + 2 Cl - (aq).
2 AgCl + Hg (l) 
o = -2 x 96500 x 0.058 = -11.194 kJ for the reaction given above. 
