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An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of \(0.10 \mathrm{M} \mathrm{NaOH}\) that is saturated with \(\mathrm{Cu}(\mathrm{OH})_{2}\), what is the cell potential at \(25^{\circ} \mathrm{C} ?\left[\mathrm{For} \mathrm{Cu}(\mathrm{OH})_{2}, K_{\mathrm{sp}}\right.\) \(\left.=1.6 \times 10^{-19} .\right]\)

Short Answer

Expert verified
The cell potential at \(25^{\circ} \mathrm{C}\) for the given electrochemical cell with a standard hydrogen electrode and a copper electrode in a \(0.10 \mathrm{M} \mathrm{NaOH}\) solution saturated with \(\mathrm{Cu}(\mathrm{OH})_{2}\) is approximately \(0.485 \mathrm{V}\).

Step by step solution

01

Write the balanced half-reactions

In order to calculate cell potential, the balanced half-reactions involving each electrode should be written. For the standard hydrogen electrode (SHE), the half-reaction is: \[\mathrm{2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)}\] For the copper electrode, the half-reaction is: \[\mathrm{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)}\]
02

Calculate the concentration of Cu²⁺ ions

To calculate the cell potential, we need to find the concentration of Cu²⁺ ions in the solution. We have the equation \(\mathrm{Cu(OH)_2(s) \rightleftharpoons Cu^{2+}(aq) + 2 OH^{-}(aq)}\) and we know that Ksp = 1.6x10⁻¹⁹ and [OH⁻] = 0.10 M (from NaOH solution). We can write Ksp as: \[K_{sp} = [\mathrm{Cu^{2+}}][\mathrm{OH^{-}}]^{2}\] Replacing the known values, we get: \[\mathrm{1.6 \times 10^{-19} = [Cu^{2+}](0.10)^{2}}\] Now, we can solve for the concentration of Cu²⁺ ions: \[[\mathrm{Cu^{2+}}] = \dfrac{1.6 \times 10^{-19}}{(0.10)^{2}} = 1.6 \times 10^{-17} \mathrm{M}\]
03

Determine the cell potential using the Nernst equation

Now that we know the concentration of Cu²⁺ ions, we can use the Nernst equation to find the cell potential. The Nernst equation is given by: \[E_{cell} = E_{\circ} - \dfrac{0.0592}{n}\log Q\] Where \(E_{\circ}\) is the standard potential, n is the number of electrons transferred, and Q is the reaction quotient. For this cell, the standard potential is the difference between the standard reduction potentials of the Cu²⁺/Cu and H⁺/H₂ couple: \(E_{\circ} = E_{\circ}(\mathrm{Cu^{2+}/Cu}) - E_{\circ}(\mathrm{H^{+}/H_{2}}) = 0.34 \mathrm{V} - 0.00 \mathrm{V} = 0.34 \mathrm{V}\) As both half-reactions involve the exchange of 2 electrons, we have n=2. The reaction quotient Q can be written as: \[Q = \dfrac{[\mathrm{Cu^{2+}}]}{[\mathrm{H^{+}}]^{2}}\] Let's assume that the concentration of H⁺ ions is 1 M (neutral solution). Therefore, Q becomes: \[Q=\dfrac{1.6\times10^{-17}}{(1)^2}=1.6\times10^{-17}\] Now, we can plug in the values into the Nernst equation to find the cell potential: \[E_{cell} = 0.34 - \dfrac{0.0592}{2}\log(1.6\times10^{-17})\] \[E_{cell} \approx0.485 \mathrm{V}\] The cell potential at 25°C is approximately 0.485 V.

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Key Concepts

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

Standard Hydrogen Electrode
The standard hydrogen electrode (SHE) is a reference electrode used in electrochemistry to measure electrode potentials. It consists of a platinum electrode that is in contact with hydrogen gas at a pressure of 1 atmosphere and immersed in a solution with hydrogen ions at a concentration of 1 molar.
The half-reaction that takes place at the SHE can be written as:
  • \[\text{2H}^{+}(\text{aq}) + \text{2e}^{-} \rightarrow \text{H}_{2}(\text{g})\]
The SHE has a standard potential of 0.00 V, making it a benchmark for comparing the potential of other electrodes. Its consistent nature provides a stable basis for calculating cell potentials in electrochemical cells. This reference electrode plays a crucial role in determining the direction and magnitude of electron flow in an electrochemical reaction.
Nernst Equation
The Nernst equation is a fundamental formula used to calculate the cell potential of an electrochemical cell under non-standard conditions. It considers the effect of ion concentration on cell potential, providing a more accurate representation of real-world scenarios. The Nernst equation is given by:
  • \[E_{cell} = E_{\circ} - \dfrac{0.0592}{n}\log Q\]
Where:
  • \(E_{cell}\) = cell potential at non-standard conditions
  • \(E_{\circ}\) = standard cell potential
  • \(n\) = number of electrons exchanged
  • \(Q\) = reaction quotient
For the copper-hydrogen cell, the equation uses the known concentrations of ions to calculate the adjusted potential at 25°C, providing insight into how concentration gradients affect cell behavior. This is especially important for understanding electrochemical reactions in real solutions.
Cell Potential
Cell potential, often described as electromotive force (emf), is a measure of the energy per charge available from an electrochemical cell. It defines the voltage difference between two electrodes and can be calculated using standard reduction potentials.
In our case, the copper half-cell and the hydrogen electrode contribute to the overall cell potential:
  • \[E_{\circ}(\text{Cu}^{2+}/\text{Cu}) = 0.34 \text{ V}\]
  • \[E_{\circ}(\text{H}^{+}/\text{H}_2) = 0.00 \text{ V}\]
The net standard cell potential \(E_{\circ}\) is thus determined by these values:
  • \[E_{\circ} = 0.34 \text{ V} - 0.00 \text{ V} = 0.34 \text{ V}\]
However, the real cell potential considers ion concentrations using the Nernst equation, resulting in a value of approximately 0.485 V. This cell potential reflects how the electron flow and setup conditions impact the overall performance of the electrochemical cell.
Copper Electrode
The copper electrode in an electrochemical cell plays a vital role as a component of the redox reaction. Copper typically functions as a cathode where reduction occurs. In this scenario, copper ions in solution (\[\text{Cu}^{2+}\]) are reduced to solid copper (\[\text{Cu(s)}\]), represented by the half-reaction:
  • \[\text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{Cu}(\text{s})\]
The concentration of copper ions is determined using the solubility product, \(K_{sp}\), of copper hydroxide (\[\text{Cu(OH)}_2\]). Knowing the ion concentration is crucial to accurately compute the cell potential using the Nernst equation.
The interaction of the copper electrode with its solution environment highlights essential principles of electrochemistry, such as the relationship between ion concentration, electrode potential, and overall cell energy efficiency.

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Most popular questions from this chapter

The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing \(\mathrm{Mn}^{2+}\) ions. These ions are then oxidized to the deeply colored \(\mathrm{MnO}_{4}^{-}\) ions by periodate ion \(\left(\mathrm{IO}_{4}{ }^{-}\right)\) in acid solution. a. Complete and balance an equation describing each of the above reactions. b. Calculate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each reaction.

Consider only the species (at standard conditions) $$ \mathrm{Br}^{-}, \mathrm{Br}_{2}, \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd} $$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s)\) ?

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{~V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{~V}\)

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). The \(\mathscr{E}^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{~V}\) relative to the standard hydrogen electrode: $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{c}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}{ }^{2-} $$ a. Calculate \(\mathscr{E}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \mathrm{~mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \times 10^{-5} M\), what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{~V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\). What is \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1\), calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

The overall reaction in the lead storage battery is \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow\) \(2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) a. For the cell reaction \(\Delta H^{\circ}=-315.9 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(263.5 \mathrm{~J} / \mathrm{K}\). Calculate \(\mathscr{E}^{\circ}\) at \(-20 .{ }^{\circ} \mathrm{C}\). Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. b. Calculate \(\mathscr{E}\) at \(-20 .{ }^{\circ} \mathrm{C}\) when \(\left[\mathrm{HSO}_{4}{ }^{-}\right]=\left[\mathrm{H}^{+}\right]=4.5 M\). c. Consider your answer to Exercise 71 . Why does it seem that batteries fail more often on cold days than on warm days?

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