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The standard reduction potentials of \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) and \(\mathrm{Cu}^{2+} /\) \(\mathrm{Cu}^{+}\)are \(0.337 \mathrm{~V}\) and \(0.153 \mathrm{~V}\) respectively. The standard electrode potential of \(\mathrm{Cu}^{+} / \mathrm{Cu}\) half cell is (a) \(0.184 \mathrm{~V}\) (b) \(0.827 \mathrm{~V}\) (c) \(0.521 \mathrm{~V}\) (d) \(0.490 \mathrm{~V}\)

Short Answer

Expert verified
The standard electrode potential of \( \mathrm{Cu}^{+}/\mathrm{Cu} \) is (a) 0.184 \mathrm{~V}.

Step by step solution

01

Understand the Reduction Potential Relationships

The information given involves the standard reduction potentials of different copper ions. The reduction potential for \( \mathrm{Cu}^{2+}/\mathrm{Cu} \) is given as \( 0.337 \mathrm{~V} \) and for \( \mathrm{Cu}^{2+}/\mathrm{Cu}^{+} \) is \( 0.153 \mathrm{~V} \). You are asked to find the reduction potential for \( \mathrm{Cu}^{+}/\mathrm{Cu} \).
02

Use the Nernst Equation for Reduction Potential Difference

To find the standard electrode potential for \( \mathrm{Cu}^{+}/\mathrm{Cu} \), we can use the given potentials and the principle that the potential difference between two reduction reactions can be obtained by subtraction. Specifically, the relation we are interested in is:\[E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}^{+}) + E^\circ(\mathrm{Cu}^{+}/\mathrm{Cu})\]
03

Solve for the Unknown Reduction Potential

Rearrange the equation to solve for \( E^\circ(\mathrm{Cu}^{+}/\mathrm{Cu}) \):\[E^\circ(\mathrm{Cu}^{+}/\mathrm{Cu}) = E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) - E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}^{+})\]Substitute the given values:\[E^\circ(\mathrm{Cu}^{+}/\mathrm{Cu}) = 0.337 \mathrm{~V} - 0.153 \mathrm{~V} = 0.184 \mathrm{~V}\]
04

Confirm the Correct Answer

Compare the calculated \( 0.184 \mathrm{~V} \) with the options given. Thus, the correct option is (a) \( 0.184 \mathrm{~V} \).

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Key Concepts

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

Reduction Potential
Reduction potential is a measure of the tendency of a chemical species to gain electrons and be reduced. It's usually expressed in volts (V). In the context of copper ions, different oxidation states have specific reduction potentials which help predict the direction of electron flow in electrochemical cells.
Reduction potentials are essential in determining the feasibility of a redox reaction. A higher reduction potential signifies a stronger affinity for electrons and thus, a greater tendency to be reduced.
For example, the standard reduction potential of the Copper (\( \mathrm{Cu}^{2+}/\mathrm{Cu} \) reaction, which is \( 0.337 \mathrm{~V} \), indicates how readily it can accept electrons compared to other reactions. Likewise, the \( \mathrm{Cu}^{2+}/\mathrm{Cu}^{+} \) reaction has a reduction potential of \( 0.153 \mathrm{~V} \). These values are utilized to calculate unknown reduction potentials, as seen in the exercise.
Nernst Equation
The Nernst Equation provides a relationship between the reduction potential of a chemical reaction and its concentration of reactants and products. It can be used to calculate the cell potential under non-standard conditions as well. The standard form of the Nernst Equation is:\[E = E^\circ - \frac{RT}{nF} \ln Q\]Where:
  • \( E \) is the cell potential at non-standard conditions,
  • \( E^\circ \) is the standard reduction potential,
  • \( R \) is the universal gas constant \((8.314 \text{ J mol}^{-1} \text{ K}^{-1})\),
  • \( T \) is the temperature in Kelvin,
  • \( n \) is the number of moles of electrons transferred in the reaction,
  • \( F \) is the Faraday constant \((96485 \text{ C mol}^{-1})\),
  • \( Q \) is the reaction quotient.
Though in this specific exercise the Nernst Equation wasn't required to find \( E^\circ(\mathrm{Cu}^{+}/\mathrm{Cu}) \), its principle of using reduction potentials to solve such problems was demonstrated. The equation helps validate calculations and understand the electrochemical behavior of cells.
Copper Ion Reactions
Copper ions exist in several oxidation states, primarily \( \mathrm{Cu}^{2+} \), \( \mathrm{Cu}^{+} \), and \( \mathrm{Cu} \). These states are pivotal in various redox reactions and are important in contexts like electroplating and energy storage technologies.
In the standard conditions, each copper ion pair has a characteristic reduction potential:
  • \( \mathrm{Cu}^{2+}/\mathrm{Cu} \) is \( 0.337 \mathrm{~V} \)
  • \( \mathrm{Cu}^{2+}/\mathrm{Cu}^{+} \) is \( 0.153 \mathrm{~V} \)
These potentials are used to predict how copper ions will interact with each other and other substances. For example, by assuming standard conditions, when two reduction potentials are given, like in the exercise, one can calculate unknown potentials by subtraction. Understanding these interactions allows scientists and industries to better harness copper’s properties for applications ranging from electrical conduction to catalysis.

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Most popular questions from this chapter

A solution of \(\mathrm{CuSO}_{4}\) is electrolyzed for 7 minutes with a current of \(0.6 \mathrm{~A}\). The amount of electricity passed equal to (a) \(26 \mathrm{C}\) (b) \(4.2 \mathrm{C}\) (c) \(2.6 \times 10^{-4} \mathrm{~F}\) (d) \(2.6 \times 10^{-3} \mathrm{~F}\)

The standard emf of a cell, involving one electron change is found to be \(0.591 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction is \(\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}\right.\) \(\left.=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(1.0 \times 10^{30}\) (b) \(1.0 \times 10^{1}\) (c) \(1.0 \times 10^{5}\) (d) \(1.0 \times 10^{10}\)

\(4.5 \mathrm{~g}\) of aluminium (at. mass \(27 \mathrm{amu}\) ) is deposited at cathode from \(\mathrm{Al}^{3+}\) solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from \(\mathrm{H}^{+}\)ions is solution by the same quantity of electric charge will be (a) \(44.8 \mathrm{~L}\) (b) \(22.4 \mathrm{~L}\) (c) \(11.2 \mathrm{~L}\) (d) \(5.6 \mathrm{~L}\)

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, \(E_{\text {cell }}^{\circ}\) is \(1.10\) volt. \(E_{\text {cell }}\) 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) \(0.82 \mathrm{~V}\)

The \(E\left(\mathrm{M}^{3+} / \mathrm{M}^{2+}\right)\) values for \(\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}\) and \(\mathrm{Co}\) are \(-0.41,+1.57,+0.77\) and \(+1.97 \mathrm{~V}\) respectively. For which one of these metals the change in oxidation state form \(+2\) to \(+3\) is easiest? (a) \(\mathrm{Cr}\) (b) \(\mathrm{Mn}\) (c) \(\mathrm{Fe}\) (d) \(\mathrm{Co}\)

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