PHYSICAL CHEMISTRY IN BRIEF

CHAP. 11: ELECTROCHEMISTRY [CONTENTS] 409

Example

Consider the following galvanic cell

Pt—H 2 (p)|HCl(a 1 )|AgCl(s)|Ag—Ag|AgCl(s)|HCl(a 2 )|Pt—H 2 (p),

wherea 2 > a 1.

Solution

Its difference from a cell with transference is that instead of a frit it has a silver chloride electrode.

It acts as a cathode with respect to the hydrogen electrode on the left of the cell record and as

an anode with respect to the hydrogen electrode on the right. The following partial reactions

proceed in the cell:

1

2 H^2 (g) = H

+(a 1 ) + e− oxidation at the hydrogen anode,

AgCl(s) + e− = Ag(s) + Cl−(a 1 ) reduction at the silver chloride electrode,

Ag(s) + Cl−(a 2 ) = AgCl(s) + e− oxidation at the silver chloride electrode,

H+(a 2 ) + e− =^12 H 2 (g) reduction at the hydrogen cathode.

E= 2

RT

F

ln

m 2 γ±, 2

m 1 γ±, 1

.

11.8.14 Gas electrode concentration cells

The cell is comprised of two identical gas electrodes containing gas of different partial pressures.
The electrodes are dipped into a shared electrolyte.

Example

Let us consider a cell formed by two hydrogen electrodes with the partial pressures of hydrogen

p 1 andp 2 , withp 1 > p 2 , dipped into a shared electrolyte.

Pt—H 2 (p 1 )|HCl(a)|Pt—H 2 (p 2 ).