For the redox reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \longrightarrow \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}(\mathrm{s})\) Taking place in a cell, \(\mathrm{E}_{\text {cell }}^{\circ}\) is \(1.10\) volt. \(\mathrm{E}_{\text {cell }}\) for the cell will be ( \(2.303 \mathrm{RT} / F=0.0591\) ): (a) \(2.14 \mathrm{~V}\) (b) \(1.80 \mathrm{~V}\) (c) \(1.07 \mathrm{~V}\) (d) \(0.82 \mathrm{~V}\)

Short Answer

Expert verified
The cell potential is approximately \(1.10 \text{ V} - 0.02955 \text{ V} = 1.07045 \text{ V}\), which is closest to option (c) \(1.07 \text{ V}\).

Step by step solution

01

Write the Nernst Equation

The Nernst equation relates the cell potential at non-standard conditions to the standard cell potential, concentration of reactants, and concentration of products. The equation is given by: \[ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \] where \( n \) is the number of moles of electrons transferred in the reaction, \( [\text{Products}] \) is the concentration of the products, and \( [\text{Reactants}] \) is the concentration of the reactants.
02

Identify the number of electrons transferred

In this redox reaction, zinc loses two electrons and copper ion gains two electrons. Thus, the number of moles of electrons transferred \( n \) is 2.
03

Plug values into the Nernst Equation

Using the given concentrations for Cu^2+ and Zn^2+, the standard cell potential \( E_{\text{cell}}^\circ \) , and \( n = 2 \), we plug these into the Nernst equation: \[ E_{\text{cell}} = 1.10 \text{ V} - \frac{0.0591}{2} \log \frac{1 \mathrm{M}}{0.1 \mathrm{M}} \]
04

Calculate the log term

The concentration ratio of Zn^2+ to Cu^2+ is \( \frac{1 \mathrm{M}}{0.1 \mathrm{M}} = 10 \). So, the logarithm of 10 is 1. Therefore, the log term is \[ \frac{0.0591}{2} \times 1 = 0.02955 \text{ V} \]
05

Finish the calculation

Subtract the log term from the standard cell potential: \[ E_{\text{cell}} = 1.10 \text{ V} - 0.02955 \text{ V} \] thus giving us the cell potential under the given conditions.

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Key Concepts

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

Redox Reaction
Redox reactions are the backbone of electrochemical cells and involve the transfer of electrons from one substance to another. These reactions consist of two half-reactions: oxidation and reduction. In oxidation, an atom or molecule loses electrons, while in reduction, an atom or molecule gains electrons.

For the given reaction, \[\mathrm{Zn(\mathrm{s}) + Cu^{2+}(0.1\ M) \longrightarrow Zn^{2+}(1\ M) + Cu(\mathrm{s})}\],the zinc (Zn) solid is being oxidized to zinc ions (Zn\textsuperscript{2+}), losing two electrons in the process. Conversely, copper ions (Cu\textsuperscript{2+}) are reduced to copper solid (Cu), gaining two electrons. The seamless flow of these electrons from zinc to copper drives the reaction forward and is key to the cell's ability to do work.
Cell Potential
The cell potential, also known as electromotive force (emf), is the measure of the driving force behind the electrical current in an electrochemical cell. It represents the energy per unit charge available from the redox reaction occurring in the cell. The cell potential is composed of potentials from each half-reaction and can be determined using standard cell potentials under standard conditions or calculated for non-standard conditions using the Nernst equation.

In the exercise, the Nernst equation is used to calculate the cell potential under non-standard conditions (different concentrations of reactants and products). The standard cell potential (\(E_{\text{cell}}^\circ\)) is a starting point for these calculations and provides the value of the potential if all reactants and products were at standard state conditions, usually 1M concentration and 1 atm pressure.
Electrochemistry
Electrochemistry is the branch of science that deals with the relationship between electrical energy and chemical reactions. It involves the study of redox reactions that occur in electrochemical cells, where chemical energy is converted into electrical energy and vice versa.

The principles of electrochemistry are applied when using the Nernst equation to determine the cell potential. The equation takes into account the innate connection between the thermodynamic properties of a chemical system (through its standard cell potential and concentrations of reactants and products) and its electrical characteristics (cell potential) to predict the behavior of an electrochemical cell.
Galvanic Cells
Galvanic cells, or voltaic cells, are a type of electrochemical cell that generates electrical energy from spontaneous redox reactions. They have two separate half-cells for the oxidation and reduction reactions, connected by an electrolyte which allows ions to move between half-cells, completed by a conductive material through which electrons can flow.

Each half-cell has its own electrode and electrolyte where half-reactions occur. The standard cell potential is the driving force of the galvanic cell under standard conditions, but real-life scenarios often deviate from these conditions, requiring the use of the Nernst equation to accurately determine the actual cell potential and therefore the electrical output of the cell.

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Most popular questions from this chapter

The emf of the following three galvanic cells: (1) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}\) (2) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}\) (3) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.1 \mathrm{M})\right| \mathrm{Cu}\) are represented by \(\mathrm{E}_{1}, \mathrm{E}_{2}\) and \(\mathrm{E}_{3} .\) Which of the following statement is true? (a) \(\mathrm{E}_{2}>\mathrm{E}_{1}>\mathrm{E}_{3}\) (b) \(\mathrm{E}_{3}>\mathrm{E}_{2}>\mathrm{E}_{1}\) (c) \(\mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}\) (d) \(\mathrm{E}_{3}>\mathrm{E}_{1}>\mathrm{E}_{2}\)

The conductivity of \(0.01 \mathrm{~mol} / \mathrm{dm}^{3}\) aqueous acetic acid at \(300 \mathrm{~K}\) is \(19.5 \times 10^{-5} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) and limiting molar conductivity of acetic acid at the same temperature is \(390 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\). The degree of dissociation of acetic acid is: (a) \(0.05\) (b) \(0.5 \times 10^{-2}\) (c) \(5 \times 10^{-7}\) (d) \(5 \times 10^{-3}\)

For the equation: \(\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+\mathrm{e} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{NO} .\) The number of electrons in its balanced form would be: (a) 6 (b) 4 (c) 3 (d) 9

The standard oxidation potential \(\mathrm{E}^{\circ}\) for the half reactions are as: \(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} ; \mathrm{E}^{\circ}=+0.76 \mathrm{~V}\) \(\mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} ; \mathrm{E}^{\circ}=+0.41 \mathrm{~V}\) The emf for the cell reaction \(\mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}\) is: (a) \(+1.17 \mathrm{~V}\) (b) \(-0.35 \mathrm{~V}\) (c) \(+0.35 \mathrm{~V}\) (d) \(0.117 \mathrm{~V}\)

Electrolysis of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) will give: (a) \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) (b) \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) (c) \(\mathrm{H}_{2} \overline{\mathrm{O}}\) and \(\mathrm{H}_{2}\) (d) \(\mathrm{SO}_{2}\) and \(\mathrm{H}_{2}\)

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