Physical Chemistry Third Edition

8.2 Electrochemical Cells 359

EXAMPLE 8.1

Find the potential difference of our cell at 298.15 K if the hydrogen pressure is equal to

744 torr and ifm 0 .500 mol kg−^1. The value ofE◦is 0.2223 V, and at this molality the

mean ionic activity, coefficient is equal to 0.757.

Solution

E 0 .2223 V−

(

(8.3145 J K−^1 mol−^1 )(298.15 K)

(2)(96485 C mol−^1 )

)

ln

(

(0.757)^4 (0.500)^4

(744 torr)/(750 torr)

)

 0 .2223 V−(− 0 .0498 V) 0 .2721 V

Our discussion of the Nernst equation applies to a particular cell. Consider now a

general cell without liquid junction. The chemical reaction equation can be written:

0 

∑c

i 1

viFi+ne−(R)−ne−(L) (8.2-18)

wherenis the number of electrons in the reaction equation. The Nernst equation for a

general cell is

EE◦−

RT

nF

ln(Q) (general Nernst equation) (8.2-19)

where

Q

∏c

i 1

avii (8.2-20)

The activity quotientQdoes not include the activity of the electron.

EXAMPLE 8.2

Write the cell symbol, the cell reaction equation, and the Nernst equation for the cell with the

half-reactions

2Hg(l)+2Cl−−→Hg 2 Cl 2 (s)+ 2 e−

Cl 2 (g)+ 2 e−−→2Cl−

Solution

The cell symbol is

Pt|Hg(l)|Hg 2 Cl 2 (s)|HCl|Cl 2 (g)|Pt

The cell reaction equation is

2Hg(l)+Cl 2 (g)+2Cl−−→Hg 2 Cl 2 (s)+2Cl−

If we omit the chloride ion from both sides, this is the same as

2Hg(l)+Cl 2 (g)−→Hg 2 Cl 2 (s)