Question

Prelab GALVANIC CELLS: THE NERNST EQUATION Cl-1 + MnO4-1  Mn+2 + Cl2 1. Balance the equation using the ion-electron (half-cell) method. Assume acidic solution. 2. Calculate the Eocell . 3. Calculate Ecell when Mn+2 and Cl-1 are each 1.0 M, MnO4-1 is 1.0 x 10-4 M, Cl2 is at 1 atm pressure, and the pH is 3.00. 4. Calculate the equilibrium constant, Keq, for the reaction.

Answer 1

1. Balancing the equation,

Here Mn is getting reduced from +7 to +2 by gain of 5e-'s

and, Cl is getting oxidized from -1 to 0 by loosing 1e-'s

The half cell reaction is,

MnO4- ----> Mn2+ (reduction)

Cl- --------> Cl2 (oxidation)

Add electrons,

MnO4- + 5e- ----> Mn2+

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

Add H2O to balance O and H+ to balance H,

MnO4- + 8H+ + 5e- ----> Mn2+ + 4H2O

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

Multiply the first equation by 2 and second by 5,

2MnO4- + 16H+ + 10e- ----> 2Mn2+ + 8H2O

10Cl- --------> 5Cl2 + 10e-

Add both equations,

2MnO4- + 10Cl- + 16H+ ---> 2Mn2+ + 5Cl2 + 8H2O

Is the final balanced equation.

2. Eocell = Ecathode - Eanode = 1.507 - 1.360 = 0.147 V

3. Ecell = Eocell - 0.0592/n log[([Mn2+]^2[pCl2]^5)/([MnO4-]^2[Cl-]^5[H+]^16)

pH = 3 = -log[H+]

[H+] = 1 x 10^-3 M

[Mn2+] = 1.0 M

[Cl-] = 1.0 M

[MnO4-] = 1 x 10^-4 M

[pCl2] = 1 atm

Feed values,

Ecell = 0.147 - 0.0592/10 log{[(1)(1)]/[(1 x 10^-4)^2(1 x 10^-3)^16(1)]}

         = -0.1845 V

4. equlibrium constant, Keq = 56