For the following cells, write the cell reaction. Find the cell potential and find the equilibrium constants.
a)
Pt (s) |Fe 2+ (aq, a = 1) Fe 3+ (aq, a = 0.1) || Cl - (aq, a = 0.001) | AgCl (s), Ag (s)
b)
Ag (s) |Ag+ (aq, a = 0.01) || Zn 2+ (aq, a = 0.1) | Zn (s)
c)
Cu (s) | Cu 2+ (aq, a = 0.1) || Zn 2+ (aq, a = 0.5) | Zn (s)
d)
Pt | H 2 | H + (1m) || Cl - | Hg 2 Cl 2 | Hg (l)
Comments:
If the cell potential is negative, it implies that the reaction is not spontaneous. In such a case, exchange the electrodes such that the electrode with a greater reduction potential is to the right and complete the calculation.
In ionic solutions, the activity of cations a+ and anions a– can not be separately determined. What is determined is the mean activity a . For simplicity this has been written as a in the above problem. Use a as given above to calculate the emfs. If one molecule of an electrolyte dissociates into + positive ions and – negative ions, then n+ = n + and n– = n – where n = no of moles of electrolyte in solution. The mean ionic activity ais defined as
av = a++ a–– where = + + – = no of ions produced per molecule of the electrolyte.
For a 1:1 electrolyte such as Na Cl, a = . For a 2:1 electrolyte such as Mg Cl 2 , a = [a + a 2– ] 1/3
22.2)
For the following reactions, write down an appropriate galvanic cell, identify the electrode reaction, show the direction of flow of electrons in the external circuit and find the value of E o for the cell
a)
Zn(s) + Cl 2 (g,1 atm) Zn 2+ (1m) + 2 Cl - (1m)
b)
Hg 22+ (1m) + Fe(s) Fe 2+ (1m) + 2Hg (l)
When temperature is not explicitly specified, use 298 K (We have generally not distinguished between 298 K and 298.15K. In accurate calculations, we need to use 298.15 K).
. For simplicity this has been written as a in the above problem. Use a as given above to calculate the emfs. If one molecule of an electrolyte dissociates into 
+ positive ions and 
. For a 2:1 electrolyte such as Mg Cl 2 , a
Zn 2+ (1m) + 2 Cl - (1m) 
