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Electrochemistry

Quiz Number:  7
Course Name:  Chemistry
Course No.132
Quiz N0. : 7
Teacher:  Billy K. Cook
Teacher's Home Page:  http://faculty.icc.cc.il.us/bcook
Teacher's Email:  Bcook@icc.cc.il.us

No. of Questions= 4

Half-way there

1 In the electrolysis os a basic solution, oxygen can be produced by the half reaction

4 OH-1aro.jpg (800 bytes)O2   +  2 H2O  +  4 e-1

How many moles of O2 can be produced from a solution that was electrolyzed for 3.00 hours using a current of 8.00 amps?

a) 1.49 x 10-2
b) 4.48 x 10-2
c) 0.224
d) 0.448
e) 0.896
2 What is the value of the reaction quotient, Q, for the cell that is constructed from the two half-reactions

Zn+2  +  2 e-1aro.jpg (800 bytes)Zn             -0.76 V

Ag+1  +  1 e-1aro.jpg (800 bytes)Ag              0.80 V

when the Zn+2 concentation is 0.0100 M and the Ag+1 concentration is 1.25 M?

a) 156
b) 125
c) 1.25 x 10-2
d) 8.00 x 10-3
e) 6.40 x 10-3
3 What is the emf at 25oC for the following cell?

Cr | Cr+3 (0.010 M) || Ag+1 (0.00010 M) | Ag

Cr+3  +  3 e-1dblaro.jpg (800 bytes)Cr                 Eo = -0.74 V

Ag+1  +  1 e-1dblaro.jpg (800 bytes)Ag                Eo = 0.80 V

a) 2.09 V
b) 1.34 V
c) 0.95 V
d) 1.74 V
e) 1.50 V
4 When a solution of cesium iodide is electrolyzed, the products that would be expected are:

Cs+1(aq)  +  1 e-1aro.jpg (800 bytes)  Cs(s)                                 Eo = -2.92 V

I2(s)  +  2 e-1aro.jpg (800 bytes)  2 I-1(aq)                                   Eo = +0.54 V

2 H2O(l)images/aro.jpg (800 bytes) O2(g)  +  4 H+1(aq)   +  4 e-1             Eo = -1.23 V

2 H2O(l)  +  2 e-1aro.jpg (800 bytes) H2(g)   +  2 OH-1(aq)           Eo = -0.83 V

a) Cs(s), I2(s).
b) H2(g), OH-1(aq), Cs(s).
c) H2(g), OH-1(aq), I2(s).
d) O2(g), H+1(aq), I2(s).
e) H2(g), OH-1(aq), O2(g), H+1(aq).





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