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Chapter 19: Problem 11

Calculate the standard emf of a cell that uses the \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell reactions at \(25^{\circ} \mathrm{C} .\) Write the equation for the cell reaction that occurs under standard-state conditions.

Short Answer

Expert verified
The standard emf of the cell is 2.71V and the cell reaction under standard conditions is \(Mg + Cu^{2+} 鈫� Mg^{2+} + Cu\).

Step by step solution

01

Identify the Half-Reactions

The given half-cell reactions are Mg/Mg2+ and Cu/Cu2+. The standard reduction potentials for these reactions from the standard reduction potential table are as follows: \n\n\(Mg^{2+} + 2e^- 鈫� Mg \, E掳 = -2.37 \, V\)\n\(Cu^{2+} + 2e^- 鈫� Cu \, E掳 = +0.34 \, V\)
02

Write the Cell Reaction and Calculate the Standard Cell Potential

An oxidation-reduction (redox) reaction consists of two half-reactions, one for oxidation and one for reduction. In this case, magnesium (Mg) undergoes oxidation to form Mg2+, and copper ions (Cu2+) undergo reduction to form copper (Cu). Therefore, the half-reactions can be written as follows:\n\nOxidation: \(Mg 鈫� Mg^{2+} + 2e^-\) \nReduction: \(Cu^{2+} + 2e^- 鈫� Cu\) \n\nThese can be added to give the overall cell reaction:\n\n\(Mg + Cu^{2+} 鈫� Mg^{2+} + Cu\) \n\nThe standard cell potential (\(E^{掳}_{cell}\)) is the difference between the standard reduction potentials of the cathode and anode. Since the Cu2+/Cu half-cell acts as the cathode (where reduction occurs) and the Mg/Mg2+ half-cell acts as the anode (where oxidation occurs), \(E^{掳}_{cell}\) can be calculated as follows:\n\n\(E^{掳}_{cell} = E^{掳}_{cathode} - E^{掳}_{anode} = 0.34V - (-2.37V) = 2.71V\)
03

Final Simplified Answer

The standard emf of the cell is calculated to be 2.71V. The cell reaction that occurs under standard-state conditions is \(Mg + Cu^{2+} 鈫� Mg^{2+} + Cu\).

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Key Concepts

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

Half-cell reactions
In electrochemistry, a half-cell is part of the whole electrochemical cell. A half-cell contains a conductive electrode and a surrounding conductive electrolyte. This setup allows for the exchange of ions that participate in the redox reaction. When dealing with half-cell reactions, understanding these components is crucial. Each half-cell undergoes either an oxidation reaction, where electrons are lost, or a reduction reaction, where electrons are gained.

In the original problem, we have two half-cell reactions:
  • The magnesium (\(\mathrm{Mg}/\mathrm{Mg}^{2+}\)) half-reaction: which involves oxidation, as magnesium turns into magnesium ions by losing electrons.
  • The copper (\(\mathrm{Cu}^{2+}/\mathrm{Cu}\)) half-reaction: which involves reduction, as copper ions gain electrons to turn back into solid copper.
These separate processes are crucial building blocks for constructing the full cell reaction and determining the cell potential.
Reduction potential
Reduction potential, often given the symbol \(E^掳\), is a measure of the tendency of a chemical species to acquire electrons and be reduced. Each species has its own standard reduction potential, and these potentials can be found in a table of standard reduction potentials.

In electrochemical cells, the half-cells are connected and the species with the higher reduction potential acts as the cathode (reduction occurs here), while the one with the lower reduction potential acts as the anode (oxidation occurs here).

For our example:
  • The standard reduction potential for \(Cu^{2+} + 2e^鈭� 鈫� Cu\) is \( +0.34 \, V\).
  • The standard reduction potential for \(Mg^{2+} + 2e^鈭� 鈫� Mg\) is \( -2.37 \, V\).
The difference between these values helps determine the standard emf of the cell. In the example given, the magnesium reaction has a lower potential, meaning it's more inclined to lose electrons and act as an anode.
Oxidation-reduction reaction
An oxidation-reduction (redox) reaction is a chemical process involving the transfer of electrons between two species. This reaction is divided into two half-reactions - one that involves oxidation and the other that involves reduction. It's a fundamental concept in electrochemistry, as redox reactions are what drive the flow of electric current in these systems.

In the case provided:
  • Oxidation occurs when magnesium (\(Mg\)) is converted into magnesium ions (\(Mg^{2+}\)), releasing electrons into the external circuit.
  • Reduction occurs when copper ions (\(Cu^{2+}\)) gain electrons and form copper (\(Cu\)).
This simultaneous process of electron transfer is what powers the electrochemical cell.

In summary, the redox reaction for the entire cell we are analyzing is represented as: \(Mg + Cu^{2+} 鈫� Mg^{2+} + Cu\). The resultant movement of electrons is why current flows, generating an emf of \(2.71 \, V\) for this specific cell.

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Most popular questions from this chapter

How many faradays of electricity are required to produce (a) \(0.84 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at exactly \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) from aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution; (b) \(1.50 \mathrm{~L}\) of \(\mathrm{Cl}_{2}\) at \(750 \mathrm{mmHg}\) and \(20^{\circ} \mathrm{C}\) from molten \(\mathrm{NaCl}\); (c) \(6.0 \mathrm{~g}\) of Sn from molten \(\mathrm{SnCl}_{2} ?\)

Calculate the standard potential of the cell consisting of the \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half-cell and the SHE. What will the emf of the cell be if \(\left[\mathrm{Zn}^{2+}\right]=0.45 \mathrm{M}, \mathrm{P}_{\mathrm{H}_{2}}=2.0 \mathrm{~atm}\), and \(\left[\mathrm{H}^{+}\right]=1.8 \mathrm{M} ?\)

Steel hardware, including nuts and bolts, is often coated with a thin plating of cadmium. Explain the function of the cadmium layer.

A \(9.00 \times 10^{2}-\mathrm{mL} 0.200 \mathrm{M} \mathrm{MgI}_{2}\) was electrolyzed. As a result, hydrogen gas was generated at the cathode and iodine was formed at the anode. The volume of hydrogen collected at \(26^{\circ} \mathrm{C}\) and \(779 \mathrm{mmHg}\) was \(1.22 \times 10^{3} \mathrm{~mL}\). (a) Calculate the charge in coulombs consumed in the process. (b) How long (in min) did the electrolysis last if a current of 7.55 A was used? (c) A white precipitate was formed in the process. What was it and what was its mass in grams? Assume the volume of the solution was constant.

To remove the tarnish \(\left(\mathrm{Ag}_{2} \mathrm{~S}\right)\) on a silver spoon, a student carried out the following steps. First, she placed the spoon in a large pan filled with water so the spoon was totally immersed. Next, she added a few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolved. Finally, she placed some aluminum foil at the bottom of the pan in contact with the spoon and then heated the solution to about \(80^{\circ} \mathrm{C}\). After a few minutes, the spoon was removed and rinsed with cold water. The tarnish was gone and the spoon regained its original shiny appearance. (a) Describe with equations the electrochemical basis for the procedure. (b) Adding \(\mathrm{NaCl}\) instead of \(\mathrm{NaHCO}_{3}\) would also work because both compounds are strong electrolytes. What is the added advantage of using \(\mathrm{NaHCO}_{3}\) ? (Hint: Consider the \(\mathrm{pH}\) of the solution.) (c) What is the purpose of heating the solution? (d) Some commercial tarnish removers containing a fluid (or paste) that is a dilute \(\mathrm{HCl}\) solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two disadvantages of using this procedure compared to the one described here.
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