Half Reactions and Reduction Potential - MCAT Chemical and Physical Foundations of Biological Systems

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Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

If there is a net production of copper ions in the half cell on the left as the reaction proceeds, which of the following must be true?

Answer

If copper ions are generated as the voltaic cell functions, then the copper is being oxidized, and the silver must be reduced. Reduction and oxidation always occur together in a coupled reaction. This must also mean that the reduction potential for Ag is higher than the reduction potential for Cu.

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Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

Instead of silver, a scientist uses a strip of zinc in the opposite half cell from copper. Which of the following is true when comparing this new cell to the cell in the diagram?

The reduction potential of $Cu^{2+}$ is 0.34 volts. The reduction potential for $Zn^{2+}$ is -0.76 volts.

Answer

In the new cell, energy can still be produced, but because zinc ions have a lower reduction potential than copper ions, the copper will be reduced and the direction of electron flow will be reversed, as compared to the cell with silver in which copper was oxidized.

Compare your answer with the correct one above

Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

As the difference in the reduction potentials between two half cells increases, what happens to the Gibbs free energy of the reaction?

Answer

The reduction potential of a cell is directly related to the Gibbs free energy by the equation below.

$\Delta G = -nFE^{o}$

As the reduction potential of a cell becomes more and more positive, the Gibbs free energy value becomes more and more negative.

Compare your answer with the correct one above

Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

In the voltaic cell drawn above, what is true about the species Ag?

The reduction potential for $Cu^{2+}$ is 0.34. The reduction potential for $Ag^{+}$ 0.80.

Answer

As drawn above, the reaction must involve the oxidation of copper (lower reduction potential) and the reduction of silver (higher reduction potential). Reduction always means that a species gains electrons.

Compare your answer with the correct one above

Question

Imagine a galvanic cell which uses solid zinc and aqueous iron ions to produce a voltage.

$\small Zn(s) + Fe^{2+}(aq)\rightarrow Zn^{2+}(aq) + Fe(s)$

$\small Zn^{2+} + 2e^{-}\rightarrow Zn$ $\small \small E^ {\circ} = -0.76V$

$\small Fe^{2+} + 2e^{-} \rightarrow Fe$ $\small E^{\circ} = -0.44V$

What is the standard state cell potential for this reaction?

Answer

Keep in mind that a galvanic cell will always have a positive voltage, so you can disregard the negative options. The half reactions show the voltage that will result if the element in question is reduced; however, an oxidation-reduction reaction will always have one element oxidized and another element reduced. In the equation shown, solid zinc (Zn) is oxidized, so the voltage of its half reaction is inverted to +0.76V. Next, you add the voltage of iron's reduction, resulting in the overall voltage of the galvanic cell.

$\small 0.76V +(-0.44V) = 0.32V$

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Question

Imagine a galvanic cell which uses solid zinc and aqueous iron ions to produce a voltage.

$\small Zn(s) + Fe^{2+}(aq)\rightarrow Zn^{2+}(aq) + Fe(s)$

$\small Zn^{2+} + 2e^{-}\rightarrow Zn$ $\small \small E^ {\circ} = -0.76V$

$\small Fe^{2+} + 2e^{-} \rightarrow Fe$ $\small E^{\circ} = -0.44V$

Suppose that this galvanic cell was converted into an electrolytic cell. Which of the following statements would be true?

Answer

An electrolytic cell is best thought of as a cell that requires an external power source in order to work. The reaction will go in the opposite direction of a galvanic cell, meaning that the cell potential will also be inversed and the reaction will be non-spontaneous. As a result, cell potential would be negative in an electrolytic cell.

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Question

Imagine a galvanic cell which uses solid zinc and aqueous iron ions to produce a voltage.

$\small Zn(s) + Fe^{2+}(aq)\rightarrow Zn^{2+}(aq) + Fe(s)$

$\small Zn^{2+} + 2e^{-}\rightarrow Zn$ $\small \small E^ {\circ} = -0.76V$

$\small Fe^{2+} + 2e^{-} \rightarrow Fe$ $\small E^{\circ} = -0.44V$

Assuming standard conditions, what is the free energy for the reaction?

$\small \small F= 96,500\frac{C}{mol}$

Answer

Since the reaction is taking place under standard conditions, we can determine the free energy of the reaction by using the equation $\small \Delta G^{\circ} = -nFE^{\circ}$ .

n is the number of moles of electrons transferred in the balanced reaction, F is Faraday's constant, and Eo is the cell potential for the reaction.

$\small \Delta G^{\circ} = -(2)(96,500)(0.32V)$

$\small \small \Delta G^{\circ} = -61.8 \frac{kJ}{mol}$

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Question

Determine the cell potential of the following reaction.

$3Ca + 2Au^{3+} \rightarrow 3Ca^{2+} + 2Au$

Reduction potentials of and cations are and , respectively.

Answer

In the given reaction calcium, , is oxidized (loses electrons) and gold, , is reduced (gains electrons).

We are only given the reduction potentials. The oxidation potential of is the negative of the reduction potential: .

Recall that only the moles of electrons must balance for these reactions, therefore no multiplication of the standard potentials is needed when balancing mole atoms; thus the cell potential is the sum of the calcium oxidation potential and the gold reduction potential.

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Question

Consider the following half reactions and corresponding reduction potentials:

$\small Half Reaction...................................E^{o}(V)$

$\small \small Cl_{2(g)}+2e^{-}\rightarrow 2Cl^{-}_{(aq)}....................+1.36$

$\small Ag^{+}+e^{-}\rightarrow Ag_{(s)}...........................+0.80$

$I_{2(s)}+2e^{-}\rightarrow 2I^{-}_{(aq)}.....................+0.54$

$Pb^{2+}+2e^{-}\rightarrow Pb_{(s)}...................-0.13$

$Li^{+}+e^{-}\rightarrow Li_{(s)}........................-3.05$

Based on the above information, which substance can oxidize $I^{-}_{(aq)}$ to $I_{2(s)}$ ?

Answer

The reduction potential of $I_{2(s)}$ is $\small +0.54$ , so the corresponding oxidation potential of $I^{-}_{(aq)}$ is $\small -0.54$ .

For a substance to oxidize $I^{-}_{(aq)}$ to $I_{2(s)}$ , it must have reduction potential greater than $\small +0.54$ , so that the sum of the reduction potential of this compounds and the oxidation potential of $I^{-}_{(aq)}$ is positive.

$E^o_{oxi}+E^o_{red}>0$

$-0.54+E^o_{red}>0$

The only choice that meets this requirement is $Cl_{2(g)}$ .

$E^{o}_{Cl_{2(g)}} + E^{o}_{I^{-}_{(aq)}} = +1.36+(-0.54)=+0.82$

Compare your answer with the correct one above

Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

If there is a net production of copper ions in the half cell on the left as the reaction proceeds, which of the following must be true?

Answer

If copper ions are generated as the voltaic cell functions, then the copper is being oxidized, and the silver must be reduced. Reduction and oxidation always occur together in a coupled reaction. This must also mean that the reduction potential for Ag is higher than the reduction potential for Cu.

Compare your answer with the correct one above

Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

Instead of silver, a scientist uses a strip of zinc in the opposite half cell from copper. Which of the following is true when comparing this new cell to the cell in the diagram?

The reduction potential of $Cu^{2+}$ is 0.34 volts. The reduction potential for $Zn^{2+}$ is -0.76 volts.

Answer

In the new cell, energy can still be produced, but because zinc ions have a lower reduction potential than copper ions, the copper will be reduced and the direction of electron flow will be reversed, as compared to the cell with silver in which copper was oxidized.

Compare your answer with the correct one above

Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

As the difference in the reduction potentials between two half cells increases, what happens to the Gibbs free energy of the reaction?

Answer

The reduction potential of a cell is directly related to the Gibbs free energy by the equation below.

$\Delta G = -nFE^{o}$

As the reduction potential of a cell becomes more and more positive, the Gibbs free energy value becomes more and more negative.

Compare your answer with the correct one above

Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

In the voltaic cell drawn above, what is true about the species Ag?

The reduction potential for $Cu^{2+}$ is 0.34. The reduction potential for $Ag^{+}$ 0.80.

Answer

As drawn above, the reaction must involve the oxidation of copper (lower reduction potential) and the reduction of silver (higher reduction potential). Reduction always means that a species gains electrons.

Compare your answer with the correct one above

Question

Imagine a galvanic cell which uses solid zinc and aqueous iron ions to produce a voltage.

$\small Zn(s) + Fe^{2+}(aq)\rightarrow Zn^{2+}(aq) + Fe(s)$

$\small Zn^{2+} + 2e^{-}\rightarrow Zn$ $\small \small E^ {\circ} = -0.76V$

$\small Fe^{2+} + 2e^{-} \rightarrow Fe$ $\small E^{\circ} = -0.44V$

What is the standard state cell potential for this reaction?

Answer

Keep in mind that a galvanic cell will always have a positive voltage, so you can disregard the negative options. The half reactions show the voltage that will result if the element in question is reduced; however, an oxidation-reduction reaction will always have one element oxidized and another element reduced. In the equation shown, solid zinc (Zn) is oxidized, so the voltage of its half reaction is inverted to +0.76V. Next, you add the voltage of iron's reduction, resulting in the overall voltage of the galvanic cell.

$\small 0.76V +(-0.44V) = 0.32V$

Compare your answer with the correct one above

Question

Imagine a galvanic cell which uses solid zinc and aqueous iron ions to produce a voltage.

$\small Zn(s) + Fe^{2+}(aq)\rightarrow Zn^{2+}(aq) + Fe(s)$

$\small Zn^{2+} + 2e^{-}\rightarrow Zn$ $\small \small E^ {\circ} = -0.76V$

$\small Fe^{2+} + 2e^{-} \rightarrow Fe$ $\small E^{\circ} = -0.44V$

Suppose that this galvanic cell was converted into an electrolytic cell. Which of the following statements would be true?

Answer

An electrolytic cell is best thought of as a cell that requires an external power source in order to work. The reaction will go in the opposite direction of a galvanic cell, meaning that the cell potential will also be inversed and the reaction will be non-spontaneous. As a result, cell potential would be negative in an electrolytic cell.

Compare your answer with the correct one above

Question

Imagine a galvanic cell which uses solid zinc and aqueous iron ions to produce a voltage.

$\small Zn(s) + Fe^{2+}(aq)\rightarrow Zn^{2+}(aq) + Fe(s)$

$\small Zn^{2+} + 2e^{-}\rightarrow Zn$ $\small \small E^ {\circ} = -0.76V$

$\small Fe^{2+} + 2e^{-} \rightarrow Fe$ $\small E^{\circ} = -0.44V$

Assuming standard conditions, what is the free energy for the reaction?

$\small \small F= 96,500\frac{C}{mol}$

Answer

Since the reaction is taking place under standard conditions, we can determine the free energy of the reaction by using the equation $\small \Delta G^{\circ} = -nFE^{\circ}$ .

n is the number of moles of electrons transferred in the balanced reaction, F is Faraday's constant, and Eo is the cell potential for the reaction.

$\small \Delta G^{\circ} = -(2)(96,500)(0.32V)$

$\small \small \Delta G^{\circ} = -61.8 \frac{kJ}{mol}$

Compare your answer with the correct one above

Question

Determine the cell potential of the following reaction.

$3Ca + 2Au^{3+} \rightarrow 3Ca^{2+} + 2Au$

Reduction potentials of and cations are and , respectively.

Answer

In the given reaction calcium, , is oxidized (loses electrons) and gold, , is reduced (gains electrons).

We are only given the reduction potentials. The oxidation potential of is the negative of the reduction potential: .

Recall that only the moles of electrons must balance for these reactions, therefore no multiplication of the standard potentials is needed when balancing mole atoms; thus the cell potential is the sum of the calcium oxidation potential and the gold reduction potential.

Compare your answer with the correct one above

Question

Consider the following half reactions and corresponding reduction potentials:

$\small Half Reaction...................................E^{o}(V)$

$\small \small Cl_{2(g)}+2e^{-}\rightarrow 2Cl^{-}_{(aq)}....................+1.36$

$\small Ag^{+}+e^{-}\rightarrow Ag_{(s)}...........................+0.80$

$I_{2(s)}+2e^{-}\rightarrow 2I^{-}_{(aq)}.....................+0.54$

$Pb^{2+}+2e^{-}\rightarrow Pb_{(s)}...................-0.13$

$Li^{+}+e^{-}\rightarrow Li_{(s)}........................-3.05$

Based on the above information, which substance can oxidize $I^{-}_{(aq)}$ to $I_{2(s)}$ ?

Answer

The reduction potential of $I_{2(s)}$ is $\small +0.54$ , so the corresponding oxidation potential of $I^{-}_{(aq)}$ is $\small -0.54$ .

For a substance to oxidize $I^{-}_{(aq)}$ to $I_{2(s)}$ , it must have reduction potential greater than $\small +0.54$ , so that the sum of the reduction potential of this compounds and the oxidation potential of $I^{-}_{(aq)}$ is positive.

$E^o_{oxi}+E^o_{red}>0$

$-0.54+E^o_{red}>0$

The only choice that meets this requirement is $Cl_{2(g)}$ .

$E^{o}_{Cl_{2(g)}} + E^{o}_{I^{-}_{(aq)}} = +1.36+(-0.54)=+0.82$

Compare your answer with the correct one above

Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

If there is a net production of copper ions in the half cell on the left as the reaction proceeds, which of the following must be true?

Answer

If copper ions are generated as the voltaic cell functions, then the copper is being oxidized, and the silver must be reduced. Reduction and oxidation always occur together in a coupled reaction. This must also mean that the reduction potential for Ag is higher than the reduction potential for Cu.

Compare your answer with the correct one above

Question

A student conducts an experiment for a chemistry class. The student wishes to explore power generation from different types of voltaic cells. He sets up three different cells, and then compares the amount of energy generated from each one.

One of his cells is shown below as an example. Both remaining cells follow the same layout.

Instead of silver, a scientist uses a strip of zinc in the opposite half cell from copper. Which of the following is true when comparing this new cell to the cell in the diagram?

The reduction potential of $Cu^{2+}$ is 0.34 volts. The reduction potential for $Zn^{2+}$ is -0.76 volts.

Answer

In the new cell, energy can still be produced, but because zinc ions have a lower reduction potential than copper ions, the copper will be reduced and the direction of electron flow will be reversed, as compared to the cell with silver in which copper was oxidized.

Compare your answer with the correct one above

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