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A galvanic cell consists of a silver electrode in contact with \(346 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) solution and a magnesium electrode in contact with \(288 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) solution. (a) Calculate \(E\) for the cell at \(25^{\circ} \mathrm{C}\). (b) A current is drawn from the cell until \(1.20 \mathrm{~g}\) of silver has been deposited at the silver electrode. Calculate \(E\) for the cell at this stage of operation.

Short Answer

Expert verified
(a) Initial \( E = 3.047 \text{ V} \). (b) After deposition, \( E = 3.078 \text{ V} \).

Step by step solution

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01

Write the Cell Diagram

The galvanic cell can be described as \( \text{Mg} \mid \text{Mg}^{2+} (0.100 M) \parallel \text{Ag}^+ (0.100 M) \mid \text{Ag} \). This means magnesium will serve as the anode and silver as the cathode.
02

Determine the Standard Electrode Potentials

The standard reduction potential for the silver electrode is \( E^\circ (\text{Ag}^+/\text{Ag}) = +0.80 \text{ V} \), and for the magnesium electrode, it is \( E^\circ (\text{Mg}^{2+}/\text{Mg}) = -2.37 \text{ V} \).
03

Calculate Standard Cell Potential

Using the formula for the cell potential \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \), we find \( E^\circ_{\text{cell}} = 0.80 \text{ V} - (-2.37 \text{ V}) = 3.17 \text{ V} \).
04

Calculate Reaction Quotient (Q)

Initially, both solutions are at 0.100 M, so \( Q = \frac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2} = \frac{0.100}{(0.100)^2} = 100 \).
05

Calculate Initial Cell Potential Using Nernst Equation

The Nernst equation is \( E = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \). At \(25^\circ C\), \( E = 3.17 - \frac{0.0257}{2} \ln(100) \approx 3.047 \text{ V} \).
06

Calculate Moles of Silver Deposited

Silver deposited is \(1.20 \text{ g} / 107.87 \text{ g/mol} = 0.01112 \text{ moles} \). As every mole of Ag needs one mole of electrons, \(0.01112 \text{ moles}\) of electrons are transferred.
07

Calculate Changes in Ion Concentrations

\(\text{Ag}^+ \) ions decrease by \(0.01112 \text{ moles}\) and \(\text{Mg}^{2+} \) ions increase by \(0.00556\text{ moles}\). The new concentrations are \([ ext{Ag}^+] = \frac{0.100 \times 0.346 - 0.01112}{0.346} = 0.0679\text{ M}\) and \([ ext{Mg}^{2+}] = \frac{0.100 \times 0.288 + 0.00556}{0.288} = 0.1193 \text{ M}\).
08

Recalculate Reaction Quotient (Q)

New \( Q \) is \( \frac{0.1193}{(0.0679)^2} \approx 25.89 \).
09

Recalculate Cell Potential Using Nernst Equation

The Nernst equation gives \( E = 3.17 - \frac{0.0257}{2} \ln(25.89) \approx 3.078 \text{ V} \).

Key Concepts

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

Nernst equation
The Nernst equation is a fundamental tool in electrochemistry that relates the cell potential to the concentration of ions in the galvanic cell. It is especially useful when the system is not in standard conditions. This is common in practical scenarios where concentrations differ from 1 M.
To calculate the cell potential under non-standard conditions, the Nernst equation is used: \[ E = E^\circ_{cell} - \frac{RT}{nF} \ln Q \] - **\(E^\circ_{cell}\)** is the standard cell potential. - **\(R\)** is the universal gas constant (8.314 J/mol·K). - **\(T\)** is the temperature in Kelvin. - **\(n\)** is the number of moles of electrons exchanged in the reaction. - **\(F\)** is Faraday's constant (96485 C/mol). - **\(Q\)** is the reaction quotient. Understanding this equation is crucial for determining how variations in concentration affect the voltage of your cell, allowing for the calculation of real-world conditions of a galvanic cell.
Standard electrode potential
The standard electrode potential is a constant that measures the tendency of a chemical species to be reduced, and it is measured under standard conditions: 1 M concentration, 1 atm pressure, and 25 °C (298 K). In electrochemical cells, standard electrode potentials help identify which metal will undergo oxidation and which will undergo reduction.
Typically, elements with higher standard potentials (more positive) are better at accepting electrons, making them favorable for reduction. Conversely, elements with lower potential values (more negative) tend to lose electrons easily, favoring oxidation.
In a galvanic cell, the net cell potential \(E^\circ_{cell}\) is calculated from the standard potentials of the cathode and anode using: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] This calculation provides the baseline voltage when all conditions are ideal, a vital first step before applying the Nernst equation.
Reaction quotient
The reaction quotient, Q, gives a snapshot of the concentrations of the reactants and products at a particular moment, not necessarily at equilibrium. It is expressed as the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients from the balanced equation.
In a galvanic cell involving magnesium and silver, the reaction quotient initially is calculated by: \[ Q = \frac{[Mg^{2+}]}{[Ag^+]^2} \] Initially, Q reflects the concentrations of ions before any significant reaction progresses. As the cell operates, the deposition or consumption of ions alters their concentrations, prompting a reevaluation of Q. This adjustment is a key part of predicting changes in the electrochemical potential over time using the Nernst equation.
Silver electrode
A silver electrode, often used as the cathode in electrochemical cells, has a high positive standard electrode potential of +0.80 V. This indicates a strong tendency to gain electrons, thus undergo reduction, competing effectively with other potential reductions in galvanic cells.
Silver ions \((Ag^+)\) in solution get reduced to solid silver \((Ag)\) at the electrode surface. This process results in the deposition of silver metal at the electrodes, as in the given exercise. For every mole of silver reduced, one mole of electrons is transferred, showcasing the simplicity and predictability of silver-based reactions. Understanding how silver electrodes behave is essential when dealing with cell potentials and shifts resulting from changes in concentration.
Magnesium electrode
Magnesium electrodes, commonly seen as the anode in galvanic cells, have a notably negative standard electrode potential of -2.37 V. This negative value signifies magnesium's strong propensity to lose electrons, making it effective at undergoing oxidation.
When magnesium serves as an anode, it releases electrons, destabilizing the metal lattice and forming magnesium ions \((Mg^{2+})\). The process elevates the concentration of these ions in the solution. This metal's high reactivity ensures it delivers significant driving force for electron flow within the galvanic cell.
If you're working on electrochemical experiments, recognizing the role of magnesium is crucial, as it significantly influences the overall cell voltage and facilitates understanding of the cell’s energetics and efficiency.

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Most popular questions from this chapter

For a number of years, it was not clear whether mercury(I) ions existed in solution as \(\mathrm{Hg}^{+}\) or as \(\mathrm{Hg}_{2}^{2+}\). To distinguish between these two possibilities, we could set up the following system: $$ \operatorname{Hg}(l) \mid \text { soln } \mathrm{A} \| \operatorname{soln} \mathrm{B} \mid \operatorname{Hg}(l)$$ where soln A contained 0.263 g mercury(I) nitrate per liter and soln B contained \(2.63 \mathrm{~g}\) mercury(I) nitrate per liter. If the measured emf of such a cell is \(0.0289 \mathrm{~V}\) at \(18^{\circ} \mathrm{C},\) what can you deduce about the nature of the mercury(I) ions?

What is a battery? Describe several types of batteries.

A \(9.00 \times 10^{2} \mathrm{~mL}\) amount of \(0.200 \mathrm{M} \mathrm{MgI}_{2}\) solution 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.

Consider a Daniell cell operating under non-standardstate conditions. Suppose that the cell's reaction is multiplied by 2 . What effect does this have on each of the following quantities in the Nernst equation: (a) \(E\) (b) \(E^{\circ},(\mathrm{c}) Q\) (d) \(\ln Q\), (e) \(n\) ?

The equilibrium constant for the reaction: $$\operatorname{Sr}(s)+\mathrm{Mg}^{2+}(a q) \rightleftharpoons \mathrm{Sr}^{2+}(a q)+\mathrm{Mg}(s)$$ is \(2.69 \times 10^{12}\) at \(25^{\circ} \mathrm{C}\). Calculate \(E^{\circ}\) for a cell made up of \(\mathrm{Sr} / \mathrm{Sr}^{2+}\) and \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) half-cells.

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