Electrochemistry - High School Chemistry

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Question

Consider the following electrolytic cell:

$Ni + Fe^{2+} \rightarrow Ni^{2+} + Fe$ $E^{\circ} = -0.21V$

What happens at the anode in the electrolytic cell?

Answer

It does not matter if the cell is galvanic or electrolytic; oxidation will always take place at the anode. This means that the nickel loses two electrons and is oxidized at the anode to generate nickel ions.

$Ni + Fe^{2+} \rightarrow Ni^{2+} + Fe$

Nickel ions and iron are products, and are neither oxidized nor reduced during the reaction. Iron ions are reduced at the cathode to generate the iron product.

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Question

Consider the following ionic equation for a galvanic cell:

$2Fe^{3+} + Sn^{2+} \rightarrow 2Fe^{2+} + Sn^{4+}$

Which of the following takes place at the anode?

Answer

$2Fe^{3+} + Sn^{2+} \rightarrow 2Fe^{2+} + Sn^{4+}$

Remember the acronym "OIL RIG:" Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

In the above reaction tin loses two electrons, and is oxidized in the process. Iron is reduced because each iron ion gains an electron.

Reduction: $Sn^{2+}\rightarrow Sn^{4+}+2e^-$

Oxidation: $2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}$

In an electrical cell, oxidation always takes place at the anode. As a result, tin is oxidized at the anode.

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Question

Consider the following redox net ionic reaction:

$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$

Which half reaction takes place at the cathode?

Answer

The reduction half reaction always takes place at the cathode. As a result, we are looking for the half reaction in which electrons combine with an ion in order to produce a metal product. Remember that reduction is a gain of electrons.

For the redox reaction above, copper ions are reduced in order to form copper metal. Since this is the reduction reaction, it must occur at the cathode.

Reduction: $Cu^{2+} + 2e^{-} \rightarrow Cu$

Oxidation: $Zn\rightarrow Zn^{2+}+2e^-$

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Question

Which of the following species is being produced at the cathode?

$Ni^{2+}_{(aq)} + Fe_{(s)} \rightarrow Ni_{(s)} + Fe^{2+}_{(aq)}$

Answer

Remember: AN OX and RED CAT (the ANode is the site of OXidation, and REDuction takes place at the CAThode). Also remember OIL RIG (Oxidation Is Loss of electrons and Reduction Is Gain of electrons). The question asks us which species is produced at the cathode (site of reduction). Also, remember that electrons always flow from anode to cathode, and that galvanic cells are spontaneous reactions (positive $\varepsilon^{\circ}$ ), and since electrons are negatively charged, they spontaneously flow from the anode (negative cell) to the cathode (positive cell) according to the law of attraction. Thus, we are looking for the species that gains electrons. Nickel goes from an oxidation state of to . $Ni_{(s)}$ is being produced because electrons are being transferred from the anode to the cathode. In this case, the nickel ions are gaining electrons to form $Ni_{(s)}$ and this can only happen at the cathode.

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Question

Which of the following species is being produced at the anode?

$Li^{+}_{(aq)} + K_{(s)} \rightarrow Li_{(s)} + K^{+}_{(aq)}$

Answer

Remember: AN OX and RED CAT (the ANode is the site of OXidation, and REDuction takes place at the CAThode). Also remember OIL RIG (Oxidation Is Loss of electrons and Reduction Is Gain of electrons). The question asks us which species is produced at the cathode (site of reduction). Also, remember that electrons always flow from anode to cathode, and that galvanic cells are spontaneous reactions (positive $\varepsilon^{\circ}$ ), and since electrons are negatively charged, they spontaneously flow from the anode (negative cell) to the cathode (positive cell) according to the law of attraction. $K^{+}_{(aq)}$ is being produced at the anode because electrons are being transferred from the anode to the cathode. In this case, the lithium ions are gaining electrons to form solid lithium at the cathode. Solid potassium is losing electrons at the anode to form $K^+_{(aq)}$ .

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Question

$Hg^{2+}_{(aq)} + 2e^- \rightarrow Hg_{(l)}$ $\varepsilon ^o = 0.85$

$Cu^{+}_{(aq)} +e^-\rightarrow Cu_{(s)}$ $\varepsilon ^o = 0.34$

$2H^{+}_{(aq)} + 2e^- \rightarrow H_2_{(g)}$ $\varepsilon ^o = 0$

$Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}$ $\varepsilon ^o = -0.76$

$Cr^{3+}_{(aq)} + 3e^- \rightarrow Cr_{(s)}$ $\varepsilon ^o = -0.74$

Which of the following species would mostly likely be reduced, if placed in a galvanic cell with another species?

Answer

Remember that reduced means to gain electron, while oxidized means to lose electrons.

Using the equation: $\varepsilon^o_{rxn}=\varepsilon^o_{products}-\varepsilon^o_{reactants}$ , for a spontaneous reaction to occur, $\varepsilon^o_{rxn}$ must be positive. With solid mercury as the product, any other solid can act as the reactants, and still give a positive $\varepsilon^o_{rxn}$ , because it has the highest $\varepsilon^o_{products}$ value.

As a result, the equation, $Hg^{2+}_{(aq)} + 2e^-\rightarrow Hg_{(l)}$ , remains unchanged, and mercury ions will gain electrons and be reduced to liquid mercury. Any other paired equation must be inverted to give electrons.

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Question

$Li^{+}_{(aq)} + e^- \rightarrow Li_{(s)}$ $\varepsilon ^o = -3.04$

$Mg^{2+}_{(aq)} +2e^- \rightarrow Mg_{(s)}$ $\varepsilon ^o= -2.38$

$Ni^{2+}_{(aq)}+ 2e^- \rightarrow Ni_{(s)}$ $\varepsilon ^o = -0.23$

$Ag^{2+}_{(aq)} + 2e^- \rightarrow Ag_{(s)}$ $\varepsilon ^o = 0.80$

$Cl_2_{(g)}+ 2e^- \rightarrow 2Cl^-_{(aq)}$ $\varepsilon ^o= 1.36$

Which of the following species would mostly likely be reduced, if placed in a electrochemical cell with another species?

Answer

Using the equation: $\varepsilon^o_{rxn}=\varepsilon^o_{products}-\varepsilon^o_{reactants}$ , for a spontaneous reaction to occur, $\varepsilon^o_{rxn}$ must be positive. With chlorine ions as the product, any other solid can act as the reactants, and still give a positive $\varepsilon^o_{rxn}$ , because it has the highest $\varepsilon^o_{products}$ value. As a result, the equation, $Cl_2_{(g)} + 2e^-\rightarrow 2Cl^-_{(aq)}$ , remains unchanged, and chlorine gas will gain electrons and reduce to chlorine ions. Any other paired equation must be inverted to give electrons.

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Question

$Li^{+}_{(aq)} + e^- \rightarrow Li_{(s)}$ $\varepsilon ^o = -3.04$

$Mg^{2+}_{(aq)} +2e^- \rightarrow Mg_{(s)}$ $\varepsilon ^o= -2.38$

$Ni^{2+}_{(aq)}+ 2e^- \rightarrow Ni_{(s)}$ $\varepsilon ^o = -0.23$

$Ag^{2+}_{(aq)} + 2e^- \rightarrow Ag_{(s)}$ $\varepsilon ^o = 0.80$

$Cl_2_{(g)}+ 2e^- \rightarrow 2Cl^-_{(aq)}$ $\varepsilon ^o= 1.36$

Which of the following species would mostly likely be oxidized, if placed in a electrochemical cell with another species?

Answer

Using the equation: $\varepsilon^o_{rxn}=\varepsilon^o_{products}-\varepsilon^o_{reactants}$ , for a spontaneous reaction to occur, $\varepsilon^o_{rxn}$ must be positive. With solid lithium as the reactant, any other solid can act as the product, and still give a positive $\varepsilon^o_{rxn}$ , because the $\varepsilon^o_{reactants}$ is the lowest value for lithium equation. Subtracting a negative number will give a positive value.

As a result, the equation, $Li^{+}_{(aq)} + e^- \rightarrow Li_{(s)}$ , will become inverted to make the solid lithium a reactant. $Li_{(s)} \rightarrow Li^{+}_{(aq)} + e^-$ . Solid lithium will give electrons, and oxidize, to reduce other ions.

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Question

$2Li^{+}_{(aq)} + Mg_{(s)} \rightarrow 2Li_{(s)} + Mg^{2+}_{(aq)}$

Which of the following species is being produced at the cathode?

Answer

Remember: AN OX and RED CAT (the ANode is the site of OXidation, and REDuction takes place at the CAThode). Also remember OIL RIG (Oxidation Is Loss of electrons and Reduction Is Gain of electrons). The question asks us which species is produced at the cathode (site of reduction). Also, remember that electrons always flow from anode to cathode.

No knowledge about whether the cell is galvanic or electrolytic is needed, because the question assumes that the chemical reaction takes places.

$Li^{+}_{(aq)}$ is being produced at the cathode because electrons transfer from the anode to the cathode. In this reaction, $Li^{+}_{(aq)}$ are gaining electrons to form $Li_{(s)}$ , and the gaining of electrons only happens at the cathode. Therefore, $Li_{(s)}$ is produced at the cathode. On the other hand, $Mg_{(s)}$ loses electrons to produce $Mg^{2+}_{(aq)}$ in the reaction, and the losing of the electrons only occur at the anode.

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Question

$2Li^{+}_{(aq)} + Mg_{(s)} \rightarrow2Li_{(s)} + Mg^{2+}_{(aq)}$

Which of the following species is being produced at the anode?

Answer

Remember: AN OX and RED CAT (the ANode is the site of OXidation, and REDuction takes place at the CAThode). Also remember OIL RIG (Oxidation Is Loss of electrons and Reduction Is Gain of electrons). The question asks us which species is produced at the anode (site of oxidation). Also, remember that electrons always flow from anode to cathode.

No knowledge about whether the cell is galvanic or electrolytic is needed, because the question assumes that the chemical reaction takes places.

$Li^{+}_{(aq)}$ is being produced at the cathode because electrons transfer from the anode to the cathode. In this reaction, $Li^{+}_{(aq)}$ are gaining electrons to form $Li_{(s)}$ , and the gaining of electrons only happens at the cathode. Therefore, $Li_{(s)}$ is produced at the cathode. On the other hand, $Mg_{(s)}$ loses electrons to produce $Mg^{2+}_{(aq)}$ in the reaction, and the losing of the electrons only occur at the anode. Therefore, $Mg^{2+}_{(aq)}$ is produced at the anode.

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Question

Toward which pole do the electrons travel in an electrolytic cell?

Answer

Reduction always occurs at the cathode, and oxidation always occurs at the anode. Since reduction is the addition of electrons, electrons must travel toward the site of reduction.

In an electrolytic cell the negative charge is on the cathode, while the positive charge is on the anode. Since an electrolytic cell requires energy to perpetuate the reaction, we are pushing the electrons against their potential gradient. The electrons, which are negatively charged, are traveling towards the cathode, which is also negatively charged.

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Question

How does an electrolytic cell differ from a galvanic cell?

Answer

Oxidation always takes place at the anode, regardless of the electrical cell type. The charges on the anode and cathode are reversed between galvanic and electrolytic cells. In electrolytic cells, the cathodes are marked negative and the anodes are marked positive. In galvanic cells, the reverse is true: cathodes are marked positive and anodes are marked negative.

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Question

$Mg^{2+}_{(aq)} + 2e^- \rightarrow Mg_{(s)}$ $\varepsilon ^o = -2.76V$

$Ca^{2+}_{(aq)} + 2e^- \rightarrow Ca_{(s)}$ $\varepsilon ^o = -2.38V$

The following reaction below takes place in an electrochemical cell:

$Mg^{2+}_{(aq)} + Ca_{(s)} \rightarrow Mg_{(s)} + Ca^{2+}_{(aq)}$

Which of the following best describes this cell?

Answer

$Mg^{2+}_{(aq)} + 2e^- \rightarrow Mg_{(s)}$ $\varepsilon ^o = -2.76V$

$Ca^{2+}_{(aq)} + 2e^- \rightarrow Ca_{(s)}$ $\varepsilon ^o = -2.38V$

Since our overall reaction includes calcium solid in the reactants, we must invert the second equation, including the sign of the electrical potential.

$Ca_{(s)} \rightarrow Ca^{2+}_{(aq)} + 2e^-$ $\varepsilon^o=2.38V$

$\varepsilon^o_{rxn}=\varepsilon^o_{products}-\varepsilon^o_{reactants}$

$\varepsilon ^o_{rxn} = -2.76+2.38 = -0.38V$

This cell is electrolytic because the electrical potential, $\varepsilon^o_{rxn}$ , is negative. Electrolytic cells involve nonspontaneous reactions, and therefore, must have an external voltage source such as a battery to drive their reactions.

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Question

$Mg^{2+}_{(aq)} + 2e^- \rightarrow Mg_{(s)}$ $\varepsilon ^o = -2.76$

$Ca^{2+}_{(aq)} + 2e^- \rightarrow Ca_{(s)}$ $\varepsilon ^o = -2.38$

The following reaction below takes place in an electrochemical cell:

$Mg_{(s)} + Ca^{2+}_{(aq)} \rightarrow Mg^{2+}_{(aq)} + Ca_{(s)}$

Is the cell galvanic or voltaic?

Answer

$Mg^{2+}_{(aq)} + 2e^- \rightarrow Mg_{(s)}$

$Ca^{2+}_{(aq)} + 2e^- \rightarrow Ca_{(s)}$

Since our overall reaction includes magnesium solid in the reactants, we must invert the first equation, including the sign of the electrical potential.

$Mg_{(s)} \rightarrow Mg^{2+}_{(aq)} + 2e^-$ $\varepsilon^o=2.76V$

$\varepsilon^o_{rxn}=\varepsilon^o_{products}-\varepsilon^o_{reactants}$

$\varepsilon ^o_{rxn} = -2.38+2.76 = 0.38V$

This cell is galvanic because the electrical potential, $\varepsilon^o_{rxn}$ , is positive. Galvanic cells involve spontaneous reactions, and therefore, do not need any external energy source to drive the reaction.

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Question

$Hg^{2+}_{(aq)} + 2e^- \rightarrow Hg_{(l)}$ $\varepsilon ^o = 0.85$

$Cu^{+}_{(aq)} +e^-\rightarrow Cu_{(s)}$ $\varepsilon ^o = 0.34$

$2H^{+}_{(aq)} + 2e^- \rightarrow H_2_{(g)}$ $\varepsilon ^o = 0$

$Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}$ $\varepsilon ^o = -0.76$

$Cr^{3+}_{(aq)} + 3e^- \rightarrow Cr_{(s)}$ $\varepsilon ^o = -0.74$

Which of the following species would mostly likely be oxidized, if placed in a electrochemical cell with another species?

Answer

Using the equation: $\varepsilon^o_{rxn}=\varepsilon^o_{products}-\varepsilon^o_{reactants}$ , for a spontaneous reaction to occur, $\varepsilon^o_{rxn}$ must be positive. With solid zinc as the reactant, any other solid can act as the product, and still give a positive $\varepsilon^o_{rxn}$ . This is because, subtracting a negative number will give a positive value.

As a result, the equation, $Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}$ , will become inverted to make the solid zinc a reactant. $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^-$ . Solid zinc will give electrons, and oxidize, to reduce other ions.

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Question

$Li^{+}_{(aq)} + e^- \rightarrow Li_{(s)}$ $\varepsilon ^o = -3.04$

$Mg^{2+}_{(aq)} +2e^- \rightarrow Mg_{(s)}$ $\varepsilon ^o = -2.38$

For the following reaction to occur does the does the electrochemical cell voltaic or galvanic?

$2Li^{+}_{(aq)} + Mg_{(s)} \rightarrow 2Li_{(s)} + Mg^{2+}_{(aq)}$

Answer

First we must rearrange the reduction potentials so that when added together, they match the reaction that takes place in the electrochemical cell.

$Li^{+}_{(aq)} + e^- \rightarrow Li_{(s)}$

In the overall reaction, $Mg_{(s)}$ is in the reactant side, so the $Mg_{(s)}$ equation must be inverted.

$Mg_{(s)} \rightarrow Mg^{2+}_{(aq)} +2e^-$

Use the equation: $\varepsilon^o_{rxn}=\varepsilon^o_{products}-\varepsilon^o_{reactants}$ to find the $\varepsilon^o_{rxn}$ .

$Li_{(s)}$ is product, while $Mg_{(s)}$ is the reactant.

$\varepsilon^o_{rxn} = -3.04 - (-2.38) = -0.66$

The cell must be electrolytic because the $\varepsilon^o_{rxn}$ value is negative. This means, this the reaction is a non-spontaneous reaction and need a applied energy source to make the reaction take place.

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Question

How does an electrolytic cell differ from a galvanic cell?

Answer

Oxidation always takes place at the anode, regardless of the electrical cell type. The charges on the anode and cathode are reversed between galvanic and electrolytic cells. In electrolytic cells, the cathodes are marked negative and the anodes are marked positive. In galvanic cells, the reverse is true: cathodes are marked positive and anodes are marked negative.

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Question

How does an electrolytic cell differ from a galvanic cell?

Answer

Oxidation always takes place at the anode, regardless of the electrical cell type. The charges on the anode and cathode are reversed between galvanic and electrolytic cells. In electrolytic cells, the cathodes are marked negative and the anodes are marked positive. In galvanic cells, the reverse is true: cathodes are marked positive and anodes are marked negative.

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Question

Which of the following differences between galvanic cells and electrolytic cells is false?

Answer

Electrolytic cells use non-spontaneous reactions that require an external power source in order to proceed. The values between galvanic and electrolytic cells are opposite of one another. Galvanic cells have positive voltage potentials, while electrolytic voltage potentials are negative. Both types of cell, however, have oxidation occur at the cathode and reduction occur at the anode.

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Question

Which statement is true of a galvanic cell?

Answer

Galvanic cells always involve spontanous oxidation-reduction reactions. In any electrochemical cell, the electrons always move from anode to cathode. Also, the anode is always the site of oxidation, and the cathode is always the site of reduction. Since the reaction is spontaneous (net release of free energy) it drives the movement of electrons from the anode to the cathode. Remember, oxidation is loss of electrons and reduction is gain of electrons. Since oxidation always occurs at the anode, we are left with an excess of electrons, making it the negative electrode. It should makes sense that the extra electrons from the anode spontaneously travel to the cathode (positive electrode).

To help remember oxidation-reduction processes, consider the mnemonics "OIL RIG" and "An Ox, Red Cat." OIL RIG stands for "oxidation is loss, reduction is gain" in reference to electrons. An Ox, Red Cat tells us that the anode is the site of oxidation, while the cathode is the site of reduction.

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