Help with Electrolytic Cells

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AP Chemistry › Help with Electrolytic Cells

Questions 1 - 8

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

Towards the cathode

Towards the anode

Towards the positive pole

More information is needed

The electrons do not travel; only the protons travel

Explanation

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.

How does an electrolytic cell differ from a galvanic cell?

The cathode is marked negative in an electrolytic cell

Oxidation takes place at the anode in an electrolytic cell, but at the cathode in a galvanic cell

Oxidation takes place at the anode in a galvanic cell, but at the cathode in an electrolytic cell

The cell potential is positive in an electrolytic cell

Explanation

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.

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

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

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?

Galvanic

Electrolytic

Neither

Both

Explanation

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

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

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.

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

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

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?

Electrolytic

Galvanic

Neither

Both

Explanation

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

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

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.

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

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

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

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

Galvanic

Both

Electrolytic

Neither

Explanation

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.

How does an electrolytic cell differ from a galvanic cell?

The cathode is marked negative in an electrolytic cell

Oxidation takes place at the anode in an electrolytic cell, but at the cathode in a galvanic cell

Oxidation takes place at the anode in a galvanic cell, but at the cathode in an electrolytic cell

The cell potential is positive in an electrolytic cell

Explanation

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

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

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

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

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

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

$Zn_{(s)}$

$Zn^{2+}_{(aq)}$

$Hg_{(l)}$

$Cu_{(s)}$

Explanation

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.

How does an electrolytic cell differ from a galvanic cell?

The cathode is marked negative in an electrolytic cell

Oxidation takes place at the anode in an electrolytic cell, but at the cathode in a galvanic cell

Oxidation takes place at the anode in a galvanic cell, but at the cathode in an electrolytic cell

The cell potential is positive in an electrolytic cell

Explanation

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