Help with Anodes and Cathodes - AP 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

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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