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A particular voltaic cell operates on the reaction $$ \mathrm{Zn}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ giving an emf of \(0.853 \mathrm{~V}\). Calculate the maximum electrical work generated when \(20.0 \mathrm{~g}\) of zinc metal is consumed.

Short Answer

Expert verified
The maximum electrical work generated is 50527.2 Joules.

Step by step solution

01

Write Down the Reaction

The given reaction is \( \mathrm{Zn}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Zn}^{2+}(aq)+2\mathrm{Cl}^{-}(aq) \). This reaction involves the oxidation of zinc and reduction of chlorine.
02

Find Moles of Zinc

Calculate the number of moles of zinc using its atomic mass. The atomic mass of zinc is approximately \(65.38\). Use the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). Substituting the values, we get \(\text{moles of Zn} = \frac{20.0\,\text{g}}{65.38\,\text{g/mol}}\).
03

Calculate Electrical Work Formula

The electrical work done by the cell can be described by the formula \( W = -nFE \), where \( n \) is the number of moles of electrons exchanged (2 for this reaction), \( F \) is Faraday's constant (96485 C/mol), and \( E \) is the electromotive force (0.853 V).
04

Solve the Numerical Formula

From Step 2, substitute the moles of zinc (0.306 mol calculated previously), and for \( W = -nFE \), use \( n = 2 \), \( F = 96485 \) C/mol, and \( E = 0.853 \) V. Calculate \( W = -2 \times 0.306 \times 96485 \times 0.853 \).
05

Compute Maximum Electrical Work

Calculate \( W = 50527.2 \) Joules.

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Key Concepts

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

Electromotive Force (emf)
Electromotive force, or emf, is a crucial concept in understanding how batteries and voltaic cells operate. In any given electrochemical reaction, emf represents the potential difference or 'pressure' that pushes electrons through a circuit. It is the driving force that allows a voltaic cell to do work by moving electrons from the anode to the cathode.
In the context of a voltaic cell, the emf can be viewed as the chemical energy conversion into electrical energy, providing the cell's voltage. In practical terms, for the reaction where zinc reacts with chlorine, resulting in an emf of 0.853 V, this voltage is the potential energy driving the circuit.
The emf of a cell is determined by the nature of the materials involved in the reaction and their respective standard electrode potentials. Each electrochemical cell has unique emf values, facilitating the calculation and prediction of the energy changes accompanying a reaction using detailed electrochemical tables.
Electrical Work
Electrical work in an electrochemical context refers to the work done by a cell when it converts chemical energy into electrical energy. This transformation is fundamental in various applications like powering devices and facilitating chemical processes.
The formula for calculating electrical work, denoted as \( W \), is given by \( W = -nFE \). In this formula:
  • \( n \) is the number of moles of electrons transferred in the reaction. For this specific reaction:
    \( \mathrm{Zn}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Zn}^{2+}(a q)+2\mathrm{Cl}^{-}(a q) \), \( n = 2 \) as two electrons are transferred per zinc atom.
  • \( F \) is Faraday's constant. It represents the charge of one mole of electrons, a value of about 96485 Coulombs/mol.
  • \( E \) is the electromotive force obtained from the cell, which is 0.853 V in this scenario.

This equation allows you to compute the maximum electrical work the cell can perform, key in understanding both energy efficiency and capacity of voltaic cells. In computing it for 20 g of zinc, the resulting work is 50527.2 Joules, signifying the energy available from the electrochemical conversion process.
Oxidation-Reduction Reaction
The heart of a voltaic cell is its electrochemical reaction, often an oxidation-reduction (redox) process. In a redox reaction, one species is oxidized (loses electrons) while another is reduced (gains electrons).
This balance is essential for the creation of current in any cell. In the given exercise, zinc undergoes oxidation, shifting from a solid state to Zn虏鈦 ions by losing electrons. Chlorine, present as Cl鈧, undergoes reduction as it gains electrons to form Cl鈦 ions.
Redox reactions are characterized by their specific half-reactions. For zinc's oxidation, it is:
\[ \mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2e^- \]
And for chlorine's reduction:
\[ \mathrm{Cl}_{2}(g) + 2e^- \rightarrow 2\mathrm{Cl}^{-}(aq) \]
Understanding these processes is vital as they underpin the function of every voltaic or galvanic cell, influencing both the emf generated and the overall efficiency of the cell in harnessing electrical energy.

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Most popular questions from this chapter

A voltaic cell whose cell reaction is $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+\mathrm{Zn}^{2+}(a q) $$ has an emf of \(0.72 \mathrm{~V}\). What is the maximum electrical work that can be obtained from this cell per mole of iron(III) ion?

A sensitive test for bismuth(III) ion consists of shaking a solution suspected of containing the ion with a basic solution of sodium stannite, \(\mathrm{Na}_{2} \mathrm{SnO}_{2}\). A positive test consists of the formation of a black precipitate of bismuth metal. Stannite ion is oxidized by bismuth(III) ion to stannate ion, \(\mathrm{SnO}_{3}^{2-}\). Write a balanced equation for the reaction.

Consider the voltaic cell $$ \operatorname{Zn}(s)\left|\mathrm{Zn}^{2+}(a q) \| \mathrm{Cr}^{3+}(a q)\right| \operatorname{Cr}(s) $$ Write the half-cell reactions and the overall cell reaction. Make a sketch of this cell and label it. Include labels showing the anode, cathode, and direction of electron flow.

Keeping in mind that aqueous \(\mathrm{Cu}^{2+}\) is blue and aqueous \(\mathrm{Zn}^{2+}\) is colorless, predict what you would observe over a several-day period if you performed the following experiments. a. A strip of \(\mathrm{Zn}\) is placed into a beaker containing aqueous \(\mathrm{Zn}^{2+}\) b. A strip of \(\mathrm{Cu}\) is placed into a beaker containing aqueous \(\mathrm{Cu}^{2+}\) c. A strip of \(\mathrm{Zn}\) is placed into a beaker containing aqueous \(\mathrm{Cu}^{2+}\) d. A strip of \(\mathrm{Cu}\) is placed into a beaker containing aqueous \(\mathrm{Zn}^{2+}\)

A voltaic cell is constructed from the following halfcells: a magnesium electrode in magnesium sulfate solution and a nickel electrode in nickel sulfate solution. The half-reactions are $$ \begin{aligned} &\mathrm{Mg}(s) \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{e}^{-} \\ &\mathrm{Ni}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(s) \end{aligned} $$ Sketch the cell, labeling the anode and cathode (and the electrode reactions), and show the direction of electron flow and the movement of cations.

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