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

1st Problum:       Ans:   1.208 V

Solution:

The Cell potential E_cell, using the Nernst equation.

E_cell = E^o_cell - (0.0592/n) log Q ( E^o_cell is standard cell potential, Q is reaction quotient for the reaction)

From the given data, we can write the balanced redox equation

2Al + 3Fe²⁺ = 2Al³⁺ + 3Fe   (total 6e- were transferred)

Fe²⁺  + 2e^-----> Fe    E^o_red. = -0.44V              ( Standard reduction potential )

Al³⁺  + 3e^-----> Al    E^o_red. = -1.66 V              ( Standard reduction potential )

Hence E^o_cell = -0.44 - (-1.66) = 1.22 V

Reaction quotient (Q) = [Al³⁺]² / [ Fe²⁺ ]³ = [0.20]²/[0.90]³ = 0.0549

From the above equation

E_cell = E^o_cell - (0.0592/n) log Q

        =  1.22 - (0.0592/6) log (0.0549)

         = 1.2076 V

         = 1.208 V

   

2nd Problum:       Ans:   0.52 V

Solution:

From the given data, we can write the balanced redox equation

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

Ag⁺  + e^-----> Ag    E^o_red. = +0.7994V              ( Standard reduction potential )

Cl₂(g) + 2 e