Chem 17

Electrochemistry

Department of Chemical Engineering, College of Engineering

University of the Philippines, Diliman, Quezon City

Date Performed: May 15, 2013

Instructor’s Name: Maro Peña

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REFERENCES

Petrucci, R. H., Herring, F. G., Madura, J.D. and Bissonnette, C (2011). General Chemistry – Principles and Modern Applications. 10th edition. Pearson Education Inc., Canada.

APPENDICES

A. Answers to Questions

1. In the electrolysis of the KX solution:

a. At which electrode, anode or cathode is X2 generated? Is this the positive or negative terminal of the electrolytic cell?

→ X2 was generated at the anode. It is the positive terminal of the electrolytic cell.

b. What will be formed at the other electrode? Write the ionic equation involved in this electrode.

→ H2 and OH- will form in the cathode part of the cell, as illustrated in the equation below.

2 H2O(l) + 2e- → H2(g) + 2 OH-(aq)

c. Prove mathematically the assumption that the [X-] was not significantly reduced in the electrolysis.

→ [X-] = 1min×60 s1 min×0.2 C1 s×1 mol e96485 C×2 mol X-2 mol e×10.030L=4.15 x 10-3M

∴ [X-] was not significantly reduced in the electrolysis.

2. Which of the following reactions are possible sources of electrical energy? Write the cathode and anode reactions for those that are.

a. Cl2(g) + 2 Br-(aq) → 2 Cl-(aq) + Br2(g)

Cathode: Cl2(g) + 2e- → 2 Cl-(aq)

Anode: 2 Br-(aq) → 2e- + Br2(g)

b. Ag+(aq) + Cl-(aq) → AgCl(s)

The reaction above is a precipitation reaction. Therefore, it cannot be a source of electrical energy.

c. H2(g) + ½ O2(g) → H2O(l)

Cathode: H2(g) → 2 H+(aq) + 2 e-

Anode: ½ O2(g) + 2 e- → O2-(aq)

3. Given the cell

Pt|Cl2 (1atm) | Cl-(0.1M) || H+ (1M), MnO4- (1M), Mn2+ (0.1M)|Pt

for which: H+, MnO4-,Mn2+ | Pt E0red=+1.51 V

Cl-Cl2| Pt E0red=+1.36 V

a. Calculate the:

i. cell potential

E°cell = E°cathode - E°anode = 1.51V – 1.36V =...