AP Chemistry : Cell Potential Under Non-standard Conditions

Study concepts, example questions & explanations for AP Chemistry

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Example Questions

Example Question #1 : Cell Potential Under Non Standard Conditions

Consider an electrochemical cell that has the following overall reaction:

2 H+(aq) + Sn (s) -> Sn2+ (aq) + H2 (aq)

 

Which of the following changes would alter the measured cell potential?

Possible Answers:

All of the above.

Adding SnCl2 to the cathode.

Lowering the pH in the cathode.

Increase the pH in the cathode.

Increasing the H2 pressure in the cathode.

Correct answer:

All of the above.

Explanation:

All of these changes would change Q and thus change the measured cell potential.

Example Question #2 : Cell Potential Under Non Standard Conditions

For the following cell reaction:

 

2 Al (s) + 3 Mn2+ (aq) -> 2 Al3+ (aq)  + 3 Mn (s)

predict if the cell potential Ecell will be larger, or smaller than Eocell for the following conditions: 

[Al3+] = 2.0 M; [Mn2+] = 1.0 M. 

Possible Answers:

Smaller

Can not be determined

Identical

Larger

Correct answer:

Smaller

Explanation:

Altering these conditions would increase Q, and thus result in a decrease in the measured cell potential.

Example Question #4 : Cell Potential Under Non Standard Conditions

For the following cell reaction:

2 Al (s) + 3 Mn2+ (aq) -> 2 Al3+ (aq)  + 3 Mn (s)

predict if the cell potential Ecell will be larger, or smaller than Eocell for the following conditions:  [Al3+] = 1.0 M; [Mn2+] = 5.0 M. 

Possible Answers:

Can not be determined

Larger

Smaller

Identical

Correct answer:

Larger

Explanation:

At these concentrations Q will become smaller, and thus log Q will become smaller.  This will give rise to a larger cell potential.

Example Question #5 : Cell Potential Under Non Standard Conditions

What is the cell potential of the following cell:

 

Zn (s) + 2 H+(aq)  ->  Zn2+ (aq)  +  H2 (g)                                  Eo = 0.76 V

 

When the [Zn2+] = 1.0 M; PH2 = 1 atm, and the pH in the cathode is 5.2?

Possible Answers:

1.59 V

0.90 V

1.32 V

1.80 V

0.45 V

Correct answer:

0.45 V

Explanation:

Example Question #1 : Cell Potential Under Non Standard Conditions

Calculate the standard cell potential of the following reaction:

Cd(s) +  MnO2 (s) + 4 H+ (aq) +  ->  Cd2+ (aq) +  Mn2+ (aq)  + 2 H2O (l)

 

Given:

 

MnO2 (s) + 4 H+ (aq)  + 2e-  ->  Mn2+ (aq) + 2 H2O (l)            Eo = 1.23 V

Cd2+ (aq)  + 2 e-   ->  Cd (s)                                                     Eo = -0.40 V

Possible Answers:

 0.0 V

1.63 V

-1.63 V

-0.83 V

0.83 V

Correct answer:

1.63 V

Explanation:

Eocell  = Eo cathode - Eoanode

Eocell = 1.23 – (-0.40) = 1.63 V

Example Question #2 : Cell Potential Under Non Standard Conditions

Calculate the standard cell potential of the following reaction:

Zn (s) + Cu2+ (aq)  ->  Zn2+ (aq) + Cu (s)

 

Given:

Zn2+(aq)+ 2 e-->  Zn (s)                                             Eo = -0.76 V

Cu2+(aq)+ 2 e-->  Cu (s)                                             Eo = 0.34 V

Possible Answers:

0.42 V

1.10 V

0.0 V

 -1.10 V

 -0.42 V

Correct answer:

1.10 V

Explanation:

Eocell  = Eo cathode - Eoanode

Eocell = 0.34 – (-0.76) = 1.10 V

Example Question #3 : Cell Potential Under Non Standard Conditions

Calculate the standard cell potential of the following reaction:

Zn (s) + 2 Ag1+ (aq)  ->  Zn2+ (aq) + 2 Ag (s)

 

Given:

Zn2+(aq)+ 2 e-->  Zn (s)                                             Eo = -0.76 V

Ag1+(aq)+ 1 e-->  Ag (s)                                             Eo = 0.80 V

Possible Answers:

0.04 V

2.36 V

-1.56 V

-0.04 V

1.56 V

Correct answer:

1.56 V

Explanation:

Eocell  = Eo cathode - Eoanode

Eocell = 0.80 – (-0.76) = 1.56 V

Example Question #4 : Cell Potential Under Non Standard Conditions

Calculate the standard cell potential of the following reaction:

3 F2 (g) + 2 Au (s) -> 6 F- (aq) + 2 Au3+

 

Given:

F2 (g) + 2 e- -> 2 F- (aq)                                                  Eo = 2.87 V

Au3+(aq)+ 3 e--> Au (s)                                                  Eo = 1.50 V

Possible Answers:

-5.61 V

-1.37 V

 4.37 V

1.37 V

5.61 V

Correct answer:

1.37 V

Explanation:

Eocell  = Eo cathode - Eoanode

Eocell = 2.87 – (1.50) = 1.37 V

Example Question #5 : Cell Potential Under Non Standard Conditions

Determine the Ecell for the following reaction at 25 C:

Zn (s) + 2 VO2+ (aq) + 4 H+ -> 2 VO2+  (aq) +  Zn2+(aq) + 2 H2O (l)

 

Given that:

VO2+ (aq)  +  2 H+ (aq) + e- ->  VO2+(aq) + H2O (l)              Eo = 1.00 V

Zn2+ (aq)  + 2 e-->  Zn (s)                                                  Eo = -0.76 V

 

And [ VO2+] = 2.0 M; [H+] = 0.50 M; [VO2+] = 1.0 x 10-2M; [Zn2+] = 1.0 x 10-1M

Possible Answers:

1.76 V

0.95 V

2.41 V

3.71 V

1.89 V

Correct answer:

1.89 V

Explanation:

 

 

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