Cell Potential Under Non-standard Conditions
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AP Chemistry › 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.
Larger
Smaller
Identical
Can not be determined
Explanation
At these concentrations Q will become smaller, and thus log Q will become smaller. This will give rise to a larger cell potential.
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
1.63 V
0.83 V
-1.63 V
-0.83 V
0.0 V
Explanation
Eocell = Eo cathode - Eoanode
Eocell = 1.23 – (-0.40) = 1.63 V
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
-1.10 V
-0.42 V
1.10 V
0.42 V
0.0 V
Explanation
Eocell = Eo cathode - Eoanode
Eocell = 0.34 – (-0.76) = 1.10 V
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?
0.90 V
0.45 V
1.80 V
1.59 V
1.32 V
Explanation
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
1.89 V
3.71 V
0.95 V
2.41 V
1.76 V
Explanation
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
0.04 V
-0.04 V
-1.56 V
1.56 V
2.36 V
Explanation
Eocell = Eo cathode - Eoanode
Eocell = 0.80 – (-0.76) = 1.56 V
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.
Increasing the H2 pressure in the cathode.
Increase the pH in the cathode.
Adding SnCl2 to the cathode.
All of the above.
Explanation
All of these changes would change Q and thus change the measured cell potential.
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
4.37 V
-1.37 V
5.61 V
-5.61 V
1.37 V
Explanation
Eocell = Eo cathode - Eoanode
Eocell = 2.87 – (1.50) = 1.37 V
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.
Larger
Smaller
Identical
Can not be determined
Explanation
Altering these conditions would increase Q, and thus result in a decrease in the measured cell potential.