Question

calculate Ecell for each of the following cells.

Al(s)|Al³⁺( 0.20M )||Fe²⁺( 0.90M )|Fe(s)

Ag(s)|Ag⁺( 0.38M )||Cl^-( 0.094M )|Cl₂(g, 0.60atm )|Pt(s)

Answer 1

1)

given

oxidation reaction : anode

Al(s) -----> Al+3 (aq) + 3e-

reduction reaction : cathode

Fe+2 (aq) + 2e-   ---> Fe+2 (s)

the net reaction is :

2 Al (s) + 3Fe+2 (aq) ----> 2 Al+3 (aq) + 3 Fe (s)

now

Eo cell = EO cathode - EO anode

Eo cell = Eo Fe+2/Fe - Eo Al+3/Al

Eo cell = -0.44 + 1.662

Eo cell = 1.222 V

now the net reaction is

2 Al (s) + 3Fe+2 (aq) ----> 2 Al+3 (aq) + 3 Fe (s)

the reaction quotient is given by

Q= [Al+3]^2 / [Fe+2]^3

we know that

E= Eo - ( 0.0591/n) log Q

consider the net oxidation reaction :

2 Al (s) ----> 2 Al+3 + 6e-

so six electrons are transferred

so n= 6

now

E= Eo -( 0.0591/6) log [Al+3]^2 / [Fe+2]^3

using given values

E= 1.222 - ( 0.0591/6) log (0.2)^2 / (0.9)^3

E= 1.234

so

the E cell for this cell is 1.234 V

2)

the oxidation reaction : anode

Ag(s) ----> Ag+ + e-

reduction reaction : cathode

Cl2 + 2e- ----> 2Cl-

the net reaction is

2Ag (s) + Cl2 ----> 2 Ag+ (aq) + 2Cl-

now

Eo = Eo cathode - EO anode

Eo = Eo Cl2/2Cl- - Eo Ag+/Ag

Eo =    1.36 -0.80

EO = 0.56 V

now

the net reaction is

2Ag (s) + Cl2 (g) ----> 2 Ag+ (aq) + 2Cl-

reaction quotient is given by

Q = [Ag+]^2 [Cl-]^2 / pCl2

using given values

Q = (0.38)^2 (0.094)^2 / 0.6

Q = 2.126 x 10-3

now

consider the net reduction reaction :

Cl2 + 2e- ---> 2Cl-

two electrons are transferred

so n= 2

now

according to nernst equation

E = Eo - (0.0591/n) log Q

E= 0.56 - (0.0591/2) log 2.126 x 10-3

E= 0.639

so the E cell value is 0.639 V