Electrolysis

1.How many grams of Cl2 can be produced from aqueous KCl in 2.0 h with current of 30.0 A?

First, find the half reaction.

Cl2→2e-+2Cl-

To convert coulombs to moles of electrons, you must know that 96485C=1Faraday

Finally, with stochiometry, you can convert moles of Cl back to grams.

2. Given the steps and the information, we are able to work backwards to calculate the time needed to plate 3.0 kg of Cu at 50.0 A.

3.0 kg Cu x 1000 g = 3000 g x 1 mol = 47.20692368 mol x 2 e-

1 kg 63.55 g 1 mol

=94.41384736 mol e- = 96845 C = 9143509.048 C x 1 sec =

1 mol e- 50 C

182870.181 sec x 1 min x 1 hour = 51 hrs.

60 sec 60 min

Nernst Equation

1.MnO4- + 8H+ + 5e- –> Mn2+ + 4H2OEº = +1.51 V

Zn2+ + 2e- –> Zn Eº = -0.76 V

Since the reaction is spontaneous, reverse the second half reaction so you still are able to obtain a positive Eº value.

MnO4- + 8H+ + 5e- –> Mn2+ + 4H2OEº = +1.51 V

Zn–> Zn2+ + 2e-Eº = +0.76 V

Adding the two reactions will give you an Eº value of +2.27 V.

Since you must multiply the two reactions in order to get a common electron value, multiply the top half reaction by 2 and the bottom half reaction by 5 to get a total of 10 electrons transferred. This is your n value.

Since you now have your values and your given information, plug the numbers into the Nernst equation:

E = 2.27 – 0.0591 x log(1.0 x 10-3 x 1.0 x 10-2) = 2.30 V

10 (2.5 x .75)

Galvanic Cells

1. Zn(s)|Zn2+(aq)||Cu2+

2. To balance the reaction, break up the reactions into half reactions.

Cr3+ –> Cr2O72-

Cl2 –> Cl-

Balance with coefficients:

2Cr3+ –> Cr2O72-

Cl2 –> 2Cl-

Add H2O to balance out the oxygen atoms:

2Cr3+ + 7H2O –> Cr2O72-

Cl2 –> 2Cl-

Add H+ to balance out the hydrogen atoms

2Cr3+ + 7H2O –> Cr2O72- + 14H+

Cl2 –> 2Cl-

Add e- to balance out the charges:

2Cr3+ + 7H2O –> Cr2O72- + 14H+ + 6e-

2e- + Cl2 –> 2Cl-

Multiply the equations by a constant if necessary to balance the number of electrons on both sides:

2Cr3+ + 7H2O –> Cr2O72- + 14H+ + 6e-

3 x [2e- + Cl2 –> 2Cl-]

Becoming

2Cr3+ + 7H2O –> Cr2O72- + 14H+ + 6e-

6e- + 3Cl2 –> 6Cl-

Once the electrons are balanced, simplify:

2Cr3+ + 7H2O –> Cr2O72- + 14H+ + 6e-

6e- + 3Cl2 –> 6Cl-

This gives us our final answer:

2Cr3+ + 7H2O –> Cr2O72- + 14H+ + 6Cl-

Now, use this balanced equation to draw the galvanic cell

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