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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\(M \mathrm{NaOH}\) that is saturated with \(\mathrm{Cu}(\mathrm{OH})_{2},\) what is the cell potential at \(25^{\circ} \mathrm{C} ?\left[\text { For } \mathrm{Cu}(\mathrm{OH})_{2}\right.\) \(K_{\mathrm{sp}}=1.6 \times 10^{-19} . ]\)

Short Answer

Expert verified
The cell potential of the electrochemical cell consisting of a standard hydrogen electrode and a copper metal electrode at 25°C is approximately 0.3685 V.

Step by step solution

01

Write down the dissolution equation for Copper(II) hydroxide

The dissolution equation for Copper(II) hydroxide in water can be written as: \[ \mathrm{Cu}(\mathrm{OH})_{2}(s) \rightleftarrows \mathrm{Cu}^{2+}(aq) + 2 \mathrm{OH}^{-}(aq) \]
02

Find the concentration of Cu2+ ions in equilibrium

Using the given \(K_{sp}=1.6 \times 10^{-19}\), we can set up the following solubility equilibrium expression: \[K_{sp} = [\mathrm{Cu}^{2+}][\mathrm{OH}^{-}]^{2}\] Since the solution contains 0.10 M of NaOH, we know the initial concentration of OH- ions is 0.10 M. Let \(x\) be the concentration of \(\mathrm{Cu}^{2+}\) ions in equilibrium, the concentration of \(\mathrm{OH}^{-}\) will then be \(0.10+x\). So, the equilibrium expression becomes: \[1.6 \times 10^{-19} = x(0.10+x)^2\]
03

Solve for x (Concentration of Cu2+ ions)

We can approximate that \(x << 0.10\), and thus neglect \(x\) in the term \(0.10+x\). Therefore, we have: \[1.6 \times 10^{-19} = x(0.10)^2\] \[x = \frac{1.6 \times 10^{-19}}{(0.10)^2}\] \[x = 1.6 \times 10^{-17} \] So, the concentration of \(\mathrm{Cu}^{2+}\) ions in equilibrium is approximately \(1.6 \times 10^{-17} \mathrm{M}\).
04

Write down the half-cell reactions

The two half-cell reactions are: Cathode (Reduction): Cu2+ + 2e- → Cu (E° = +0.34 V) Anode (Oxidation): 2H+ + 2e- → H2 (Standard Hydrogen Electrode, E° = 0.00 V)
05

Calculate the cell potential using the Nernst equation

Now we can use the Nernst equation to determine the cell potential at 25°C (298.15 K): \[E_{cell} = E_{cathode}° - E_{anode}° - \frac{RT}{nF} \ln\frac{[\mathrm{H}^{+}]^{2}}{[\mathrm{Cu}^{2+}]}\] where R = 8.314 J/mol·K (gas constant), T = 298.15 K (temperature), n = 2 (moles of electrons exchanged), and F = 96485 C/mol (Faraday's constant). Remember that \(E_{anode}° = 0.00 \, \mathrm{V}\) since the anode is the standard hydrogen electrode. We are given \(E_{cathode}° = +0.34 \, \mathrm{V}\), and we know that pH of a 0.10 M NaOH solution is \(pH = 13\), so the concentration of \(\mathrm{H}^{+}\) ions can be calculated as \([\mathrm{H}^{+}] = 10^{-pH} = 10^{-13}\,\mathrm{M}\). Inserting these values into the Nernst equation and solving for \(E_{cell}\): \[E_{cell} = 0.34 - \frac{8.314 \cdot 298.15}{2 \cdot 96485} \ln{\frac{(10^{-13})^2}{1.6 \times 10^{-17}}}\] \[E_{cell} = 0.34 + 0.0285\] \[E_{cell} = 0.3685 \, \mathrm{V}\] Therefore, the cell potential of the electrochemical cell is approximately 0.3685 V at 25°C.

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Key Concepts

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

Nernst equation
The Nernst equation is essential in electrochemistry. It allows you to determine the potential of an electrochemical cell based on the concentrations of the reactants and products involved. The general form of the Nernst equation is:
\[E = E^0 - \frac{RT}{nF} \ln Q\]Where:
  • \(E\) is the cell potential under non-standard conditions.
  • \(E^0\) 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 transferred in the redox reaction.
  • \(F\) is Faraday's constant (96485 C/mol).
  • \(Q\) is the reaction quotient.
In the context of an electrochemical cell, using the Nernst equation helps in calculating the cell potential by considering the concentrations of ions involved in both half-reactions. This can be crucial for reactions occurring outside the standard conditions.
Using the values provided, you can adjust the potential you calculate using the Nernst equation to find the true value of cell potential for different conditions.
Standard hydrogen electrode
At the heart of many electrochemical calculations is the standard hydrogen electrode (SHE), which serves as a universal reference. It's defined under standard conditions, where its potential is universally taken to be 0.00 V. This electrode involved the half-reaction:
\[2H^+ + 2e^- \rightarrow H_2(g)\]The SHE is crucial because it offers a stable baseline for measuring electrode potentials. By coupling it with other electrodes, like the copper electrode in this exercise, you can measure the electrode potentials confidently and standardize measurements across different conditions.
Using the SHE allows scientists and engineers to compare the potentials of different half-reactions accurately. In the exercise, it's paired with the copper electrode to establish an understandable comparison for the potential calculation.
Dissolution equilibrium
Dissolution equilibrium involves the balance between a solute dissolving into a solvent and its recrystallization. In electrochemistry, this comes into play when dealing with a saturated solution and a sparingly soluble compound, like copper(II) hydroxide.
This equilibrium can be described using the solubility product constant \(K_{sp}\), shown in the reaction:
\[\mathrm{Cu}(\mathrm{OH})_2(s) \rightleftarrows \mathrm{Cu}^{2+}(aq) + 2 \mathrm{OH}^{-}(aq)\]The \(K_{sp}\) helps establish the concentrations of the ions at saturation, a critical aspect when applying principles like the Nernst equation.
Given copper(II) hydroxide's low \(K_{sp}\), the number of dissolved ions is minimal. In our scenarios, this impacts the concentration of \(\mathrm{Cu}^{2+}\) ions directly, which is a crucial step for computing the cell potential.
Cell potential calculation
Estimating an electrochemical cell's potential involves considering multiple factors, including the electrodes involved and ionic concentrations. After understanding the dissolution equilibrium and the standard potentials, you use the Nernst equation to compute the precise cell potential based on current conditions.
In the exercise, the cell contains a standard hydrogen electrode and a copper electrode. Each electrode has specific standard potentials: 0.00 V for the hydrogen electrode and specific values relevant to copper. Given the ionic concentrations from the dissolution equilibrium, you can adopt these into the Nernst equation:
\[ \large E_{cell} = E_{cathode}^0 - E_{anode}^0 - \frac{RT}{nF} \ln \frac{[H^+]^2}{[Cu^{2+}]} \]
Upon substituting the known values and conducting the calculations, the exact cell potential at 25°C in this instance is approximately 0.3685 V.
This accurate approach allows prediction of electrochemical behavior under variable conditions, providing thorough insights into how the system will perform.

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Most popular questions from this chapter

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\) if the copper electrode is placed in a solution in which \(\left[\mathrm{Cu}^{2+}\right]=\) \(2.5 \times 10^{-4} \mathrm{M} ?\) b. The copper electrode is placed in a solution of unknown \(\left[\mathrm{Cu}^{2+}\right] .\) The measured potential at \(25^{\circ} \mathrm{C}\) is 0.195 \(\mathrm{V}\) . What is \(\left[\mathrm{Cu}^{2+}\right] ?\) (Assume \(\mathrm{Cu}^{2+}\) is reduced.)

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M(\text { right side })\) and \(1.0 \times 10^{-4} M(\text { left side })\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M} .\)

Given the following two standard reduction potentials, $$\begin{array}{ll}{\mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M}} & {\mathscr{E}^{\circ}=-0.10 \mathrm{V}} \\ {\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}} & {\mathscr{E}^{\circ}=-0.50 \mathrm{V}}\end{array}$$ solve for the standard reduction potential of the half-reaction $$\mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+}$$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

Consider the following electrochemical cell: a. If silver metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. b. If copper metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. c. If the above cell is a galvanic cell, determine the standard cell potential. d. If the above cell is an electrolytic cell, determine the minimum external potential that must be applied to cause the reaction to occur.

The measurement of \(\mathrm{pH}\) using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$\mathscr{E}_{\text { meas }}=\mathscr{E}_{\text { ref }}+0.05916 \mathrm{pH}$$ where \(\mathscr{E}_{\text { ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{V}\) and that \(\mathscr{E}_{\text { meas }}=0.480 \mathrm{V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the nncertainty in the measured potential is \(+1 \mathrm{mV}\) \(( \pm 0.001 \mathrm{V}) ?\) b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

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