Cell Potential Under Non-standard Conditions - AP Chemistry

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Question

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

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 0.80 – (-0.76) = 1.56 V

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Question

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

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 2.87 – (1.50) = 1.37 V

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Question

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?

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log: Q$

$[H^+ ] = 10^{-pH}= 10^{-5.2} = 6.31 x 10^{-6}$

$E_{cell} = 0.76 V - \frac{0.0592}{2} : log : \frac{[Zn^{2+} ] P_{H_2} }{[H^+ ]^2 }$

$E_{cell} = 0.76 - \frac{0.0592}{2} : log: \frac{[1.0]1}{[6.31 x 10^{-6} ]^2} = 0.45 V$

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Question

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

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 1.23 – (-0.40) = 1.63 V

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Question

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

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 0.34 – (-0.76) = 1.10 V

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Question

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?

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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

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Question

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.

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n}: log: Q$

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

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Question

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

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Answer

$E_{cell} = E^0_{cell} - \frac {0.0592}{n} : log : Q$

$E^o_{cell} = 1.00 - (-0.76) = 1.76 V$

$E_{cell} = 1.76 V - \frac{ 0.0592}{2} : log : \frac{[VO^{2+} ]^2 [Zn^{2+} ]}{[VO_2^+ ]^2 [H^+ ]^4 )}$

$= 1.76 V - \frac{0.0592}{2}: log : \frac{[1.0 x 10^{-2} ]^2 [1.0 x 10^{-1} ]}{[2.0]^2 [0.5]^4} = 1.89$

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Question

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.

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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

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Question

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

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 0.80 – (-0.76) = 1.56 V

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Question

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

Tap to reveal answer

Answer

Eocell = Eo cathode - Eoanode

Eocell = 2.87 – (1.50) = 1.37 V

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Question

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?

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log: Q$

$[H^+ ] = 10^{-pH}= 10^{-5.2} = 6.31 x 10^{-6}$

$E_{cell} = 0.76 V - \frac{0.0592}{2} : log : \frac{[Zn^{2+} ] P_{H_2} }{[H^+ ]^2 }$

$E_{cell} = 0.76 - \frac{0.0592}{2} : log: \frac{[1.0]1}{[6.31 x 10^{-6} ]^2} = 0.45 V$

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Question

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

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 1.23 – (-0.40) = 1.63 V

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Question

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

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 0.34 – (-0.76) = 1.10 V

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Question

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?

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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

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Question

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.

Tap to reveal answer

Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n}: log: Q$

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

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Question

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

Tap to reveal answer

Answer

$E_{cell} = E^0_{cell} - \frac {0.0592}{n} : log : Q$

$E^o_{cell} = 1.00 - (-0.76) = 1.76 V$

$E_{cell} = 1.76 V - \frac{ 0.0592}{2} : log : \frac{[VO^{2+} ]^2 [Zn^{2+} ]}{[VO_2^+ ]^2 [H^+ ]^4 )}$

$= 1.76 V - \frac{0.0592}{2}: log : \frac{[1.0 x 10^{-2} ]^2 [1.0 x 10^{-1} ]}{[2.0]^2 [0.5]^4} = 1.89$

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Question

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.

Tap to reveal answer

Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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

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Question

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?

Tap to reveal answer

Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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

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Question

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.

Tap to reveal answer

Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n}: log: Q$

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

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Cell Potential Under Non-standard Conditions - AP Chemistry

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

Eocell = Eo cathode - Eoanode Eocell = 0.80 – (-0.76) = 1.56 V

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

Eocell = Eo cathode - Eoanode Eocell = 2.87 – (1.50) = 1.37 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?

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

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

Eocell = Eo cathode - Eoanode Eocell = 0.34 – (-0.76) = 1.10 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?

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

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.

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

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.

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: 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

Eocell = Eo cathode - Eoanode Eocell = 0.80 – (-0.76) = 1.56 V

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

Eocell = Eo cathode - Eoanode Eocell = 2.87 – (1.50) = 1.37 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?

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

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

Eocell = Eo cathode - Eoanode Eocell = 0.34 – (-0.76) = 1.10 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?

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

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.

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

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.

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

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?

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

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.

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

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

Eocell = Eo cathode - Eoanode

Eocell = 0.80 – (-0.76) = 1.56 V

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

Eocell = Eo cathode - Eoanode

Eocell = 2.87 – (1.50) = 1.37 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?

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

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

Eocell = Eo cathode - Eoanode

Eocell = 0.34 – (-0.76) = 1.10 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?

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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

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.

$E_{cell} = E^0_{cell} - \frac{0.0592}{n}: log: Q$

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

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.

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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:

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

Eocell = Eo cathode - Eoanode

Eocell = 0.80 – (-0.76) = 1.56 V

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

Eocell = Eo cathode - Eoanode

Eocell = 2.87 – (1.50) = 1.37 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?

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

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

Eocell = Eo cathode - Eoanode

Eocell = 0.34 – (-0.76) = 1.10 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?

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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

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.

$E_{cell} = E^0_{cell} - \frac{0.0592}{n}: log: Q$

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

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.

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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

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?

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

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

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.

$E_{cell} = E^0_{cell} - \frac{0.0592}{n}: log: Q$

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