Redox Practice Problems - Weebly



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Redox Unit Test Review

1. Define the following:

a) Oxidation:

b) Reduction:

c) Oxidizing agent:

d) Reducing agent:

2. Assign oxidation states to this reaction and determine which element is oxidized/reduced and which one is the reducing agent/oxidizing agent.

3H2S + 2HNO3 ( 3S + 2NO + 4H2O

3. Write oxidation and reduction half-reactions for this equation and balance the equation.

S(s) + HNO3(aq) ( SO2 + NO(g) + H2O(l)

4. Draw an electrochemical cell with 1.0 M CuSO4 and a Cu electrode on one side and 1.0 M ZnSO4 and a Zn electrode on the other side. Then determine:

a) Which element is more easily oxidized/reduced

b) Anode and cathode

c) Positive and negative electrodes

d) Which direction the electrons and ions are flowing

e) The reaction for which the reaction is spontaneous

f) Electric cell potential for this cell

5. Calculate the cell potential for these reactions and determine if they are spontaneous or not:

a) Mg + Sn2+ ( Sn + Mg2+

b) K + Al3+ ( K+ + Al

6. Using the Nernst equation, determine what the cell potential is for the following reaction given nonstandard conditions of [Cu2+] = 2.5 M and [Cl-] = 4.0 M:

Cu + Cl2 ( Cu2+ + 2Cl-

7. Balance the following equations with the redox half reaction method:

a) Fe+2 + Cr2O7-2 ( Fe+3 + Cr+3 (acidic solution)

b) PbO2 + I2 → Pb2+ + IO3- (acidic solution)

c) H3AsO4(aq)   +   Zn(s)   →   AsH3(g)   +   Zn2+(aq) (acidic solution)

Redox Practice Problem Answers

1. Consult book and notes.

2. 3H2S + 2HNO3 ( 3S + 2NO + 4H2O

+1 -2 +1+5-2 0 +2-2 +1-2

Sulfur is being oxidized and the reducing agent

Nitrogen is being reduced and the oxidizing agent

3. 3S(s) + 4HNO3(aq) ( 3SO2 + 4NO(g) + 2H2O(l)

4. a. Zinc is more oxidizable, Cu is more reducible

b. Zinc is the anode, Cu is the cathode

c. Cu is positive electrode, Zn is negative electrode

d. electrons flowing to Cu(s), ions flowing to CuSO4

e. Zn + Cu2+ ( Zn2+ + Cu

f. E = +.34 V + .76 V = 1.10 V

5. a. E = Ecathode + Eanode = Sn + Mg

= -.14 + 2.37 V = 2.23 V

b. E = K + Al

= -1.66 V + 2.93 V = 1.27 V

6. E = Eo - [pic] = 1.02 V -[pic]2

E = .97 V

7. be sure your half reactions are shown (only final answer shown below)

a. 6Fe+2 + Cr2O7-2 + 14H+ ( 6 Fe+3 + 2Cr+3 +7H2O

b. 5PbO2 + I2 + 8H+1 → 5Pb2+ + 2IO3- + 4 H2O

c. 8H+1 + H3AsO4(aq)   +  4Zn(s)   →   AsH3(g)   +  4 Zn2+(aq) + 4 H2O

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