AP Chemistry : Cell Potential Under Non-standard Conditions
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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?
Lowering the pH in the cathode.
All of the above.
Increasing the H2 pressure in the cathode.
Increase the pH in the cathode.
Adding SnCl2 to the cathode.
All of the above.
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.
Smaller
Larger
Identical
Can not be determined
Smaller
Altering these conditions would increase Q, and thus result in a decrease in the measured cell potential.
Example Question #1 : 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.
Identical
Can not be determined
Larger
Smaller
Larger
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 #71 : Thermodynamics
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?
1.32 V
1.80 V
0.45 V
0.90 V
1.59 V
0.45 V
Example Question #2 : 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
0.83 V
-0.83 V
-1.63 V
1.63 V
0.0 V
1.63 V
Eocell = Eo cathode - Eoanode
Eocell = 1.23 – (-0.40) = 1.63 V
Example Question #71 : Thermodynamics
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
-0.42 V
0.42 V
0.0 V
-1.10 V
1.10 V
1.10 V
Eocell = Eo cathode - Eoanode
Eocell = 0.34 – (-0.76) = 1.10 V
Example Question #72 : Thermodynamics
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
-1.56 V
2.36 V
-0.04 V
1.56 V
0.04 V
1.56 V
Eocell = Eo cathode - Eoanode
Eocell = 0.80 – (-0.76) = 1.56 V
Example Question #8 : 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
-5.61 V
4.37 V
-1.37 V
1.37 V
5.61 V
1.37 V
Eocell = Eo cathode - Eoanode
Eocell = 2.87 – (1.50) = 1.37 V
Example Question #6 : 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
3.71 V
0.95 V
1.89 V
2.41 V
1.76 V
1.89 V
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