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Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

Step by step solution

01

Identify the half reactions

It is given that copper electrode is the anode and zinc penny is the cathode. In an electrochemical cell, oxidation happens at the anode and reduction at the cathode. Thus, the two relevant reactions are: \(Cu(s) \rightarrow Cu^{2+} + 2e^-\) at the anode and \(Zn^{2+} + 2e^- \rightarrow Zn(s)\) at the cathode. Note that the reactions are reversed from the table because the copper reaction is actually an oxidation.

02

Calculate the voltage difference

The total voltage of the cell (E°cell) is equal to the difference between the reduction potentials of the cathode and the anode. So we have: E°cell = E°cathode – E°anode = (-0.76 V) - (+0.34 V) = -1.10 V.

03

Correct the sign

We've found that the cell potential is -1.10 V. But when talking about required voltages for electrochemical cells to function we always give positive values, because you can think of this as the amount of 'push' needed to make the cell run. So we correct the sign to get +1.10 V as our answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reduction Potentials

Reduction potentials are a measure of a substance's tendency to gain electrons and be reduced. These are often given in volts (V) and are a key component in electrochemical cells.

In the exercise provided, we see the use of standard reduction potentials to determine the overall cell potential. The table shows different half-reactions, each with its own reduction potential. To find the required voltage for the cell, you need to calculate the difference between the reduction potential of the cathode and the anode.

Key points to remember about reduction potentials:

• Reduction occurs at the cathode.
• Oxidation occurs at the anode.
• The cell potential (E°cell) is calculated using the formula: \(E°_{cell} = E°_{cathode} - E°_{anode}\).

This calculation helps us understand the amount of electrical energy involved in making a particular cell function. It's important that the result of this equation is *positive* for the electrochemical cell to operate spontaneously.

Electroplating

Electroplating is a process that uses an electric current to reduce dissolved metal cations, forming a coherent metal coating on an electrode. In the case of pennies, electroplating deposits copper on the surface of zinc pennies.

The setup described in the exercise involves a copper electrode (anode) and a zinc penny (cathode). In this electroplating process, copper ions are deposited onto the penny, giving it the characteristic copper coating. This transformation is facilitated by the reduction process occurring at the cathode where copper ions gain electrons and become copper metal: \(Cu^{2+} + 2e^- \rightarrow Cu(s)\).

Points to note about electroplating:

• The cathode is the site of metal deposition.
• Reduction potentials indicate how easily a metal will plate out of solution.
• The process can protect, decorate, or provide specific surface properties to an item.

This method is widely used in various industries, from jewelry manufacturing to electronics, for a range of functional and decorative purposes.

Oxidation Reactions

Oxidation reactions involve the loss of electrons by a molecule, atom, or ion. In the context of electrochemical cells, oxidation takes place at the anode, and this can be remembered with the phrase 'AnOx' (anode oxidation).

For the exercise scenario, the copper serves as the anode where the oxidation takes place. The oxidation reaction considered is \(Cu(s) \rightarrow Cu^{2+} + 2e^-\). Here, solid copper loses electrons, which travel through the external circuit to the cathode.

Essential points about oxidation reactions:

• Occurs at the anode of an electrochemical cell.
• Involves the release of electrons from the oxidized species.
• Works in tandem with reduction reactions (which occur at the cathode) to drive the cell's operation.

Understanding these reactions is crucial for gauging how energy conversion happens in these cells, linking to broader applications such as corrosion prevention, energy storage, and chemical sensing.