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High School Chemistry : Help with Galvanic Cells

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Example Questions

Example Question #1 : Electrochemistry

Which statement is true of a galvanic cell?

Possible Answers:

Electrons travel from cathode to anode, reduction occurs at the anode, the cathode is the negative electrode, and the reaction is nonspontaneous

Electrons travel from cathode to anode, oxidation occurs at the anode, the anode is the negative electrode, and the reaction is spontaneous

Electrons travel from anode to cathode, oxidation occurs at the cathode, the cathode is the negative electrode, and the reaction is spontaneous

Electrons travel from anode to cathode, reduction occurs at the cathode, the anode is the negative electrode, and the reaction is nonspontaneous

Electrons travel from anode to cathode, reduction occurs at the cathode, the anode is the negative electrode, and the reaction is spontaneous

Correct answer:

Electrons travel from anode to cathode, reduction occurs at the cathode, the anode is the negative electrode, and the reaction is spontaneous

Explanation:

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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Example Question #2 : Galvanic (Voltai) And Elecrolytic Cells

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

Possible Answers:

Electrolytic cells have a positive Gibb's free energy

Electrolytic cells are non-spontaneous

Electrolytic cells have negative voltages

Electrolytic cells have oxidation take place at the cathode

Correct answer:

Electrolytic cells have oxidation take place at the cathode

Explanation:

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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Example Question #1 : Electrochemistry

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

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

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

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

Possible Answers:

Electrolytic

Galvanic

Both

Neither

Correct answer:

Galvanic

Explanation:

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

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

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

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

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

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

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

The cell must be galvanic because the $\varepsilon^o_{rxn}$ $\displaystyle \varepsilon^o_{rxn}$ value is positive. This means, this the reaction is a spontaneous reaction occurs without an outside energy source.

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High School Chemistry : Help with Galvanic Cells

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #1 : Electrochemistry

Example Question #2 : Galvanic (Voltai) And Elecrolytic Cells

Example Question #1 : Electrochemistry

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