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An electrochemical cell is made by placing a zinc electrode in \(1.00 \mathrm{~L}\) of \(0.200 \mathrm{M} \mathrm{ZnSO}_{4}\) solution and a copper electrode in \(1.00 \mathrm{~L}\) of \(0.0100 \mathrm{M} \mathrm{CuCl}_{2}\) solution. a. What is the initial voltage of this cell when it is properly constructed? b. Calculate the final concentration of \(\mathrm{Cu}^{2+}\) in this cell if it is allowed to produce an average current of \(1.00\) amp for \(225 \mathrm{~s}\).

Short Answer

Expert verified
a. The initial voltage is approximately 1.0114 V. b. The final concentration of Cu虏鈦 is 0.008835 M.

Step by step solution

01

Identify Half-Reactions and Calculate Standard Cell Potential

Start by writing the half-reactions for the electrochemical cell. The oxidation half-reaction at the zinc electrode is:\[ \mathrm{Zn(s)} \rightarrow \mathrm{Zn}^{2+}(aq) + 2\mathrm{e}^- \] The reduction half-reaction at the copper electrode is:\[ \mathrm{Cu}^{2+}(aq) + 2\mathrm{e}^- \rightarrow \mathrm{Cu(s)} \] Using standard electrode potentials, the standard cell potential (\(E^\circ_{\text{cell}}\)) is calculated as:\[ E^\circ_{\text{cell}} = E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} - E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = 0.34 \, V - (-0.76 \, V) = 1.10 \, V \]
02

Use Nernst Equation to Find Initial Voltage

Since the concentrations are not standard, the actual initial voltage needs to be calculated using the Nernst equation:\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]Where:- \(R = 8.314 \, J/(mol \cdot K)\) is the gas constant,- \(T\) is the temperature in Kelvin (approximately 298 K for standard conditions),- \(n = 2\) is the number of moles of electrons exchanged, and- \(F = 96485 \, C/mol\) is the Faraday constant.The reaction quotient \(Q\) is given by:\[ Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]} = \frac{0.200}{0.0100} = 20 \]Plug these into the Nernst equation:\[ E_{\text{cell}} = 1.10 \, V - \frac{(8.314)(298)}{(2)(96485)} \ln(20) = 1.10 \, V - 0.0296 \, V( \ln 20) \]Calculate it to find the initial voltage.
03

Calculate the Initial Voltage

Evaluate the above expression numerically:\[ \ln(20) \approx 2.9957 \, \text{(using a calculator)}\]\[ \Rightarrow E_{\text{cell}} = 1.10 - 0.0296 \times 2.9957 \approx 1.10 - 0.0886 = 1.0114 \, V \]
04

Use Faraday's Laws to Calculate Moles of Cu Deposited

To find how much \(\mathrm{Cu}^{2+}\) is reduced, calculate the moles of electrons transferred using Faraday's second law:\[ Q = I \times t = 1.00 \, A \times 225 \, s = 225 \, C \]Now find the moles of electrons using Faraday's constant:\[ \text{Moles of electrons} = \frac{225}{96485} \approx 0.00233 \, mol \]
05

Calculate the Change in Concentration of Cu虏鈦 Ions

Since the reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) involves 2 electrons per ion, the moles of \(\mathrm{Cu}^{2+}\) reduced are half the moles of electrons:\[ \text{Moles of } \mathrm{Cu}^{2+} = \frac{0.00233}{2} = 0.001165 \, mol \]These moles will decrease the original Cu虏鈦 concentration:\[ \text{Initial moles of } \mathrm{Cu}^{2+} = 0.0100 \, mol/L \times 1.00 \, L = 0.0100 \, mol \]\[ \text{Final moles of } \mathrm{Cu}^{2+} = 0.0100 - 0.001165 = 0.008835 \]The final concentration is:\[ [\mathrm{Cu}^{2+}] = \frac{0.008835}{1.00} = 0.008835 \, M \]

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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 a critical tool when working with electrochemical cells. It allows us to calculate the cell potential at non-standard conditions. In essence, it adjusts the standard cell potential \(E^\circ_{\text{cell}}\) for changes in concentration, temperature, and pressure.

The formula is:
  • \(E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q\)
Here, \(R\) is the gas constant (8.314 J/(mol路K)), \(T\) is the temperature in Kelvin (often 298 K), \(n\) is the number of moles of electrons exchanged, and \(F\) is Faraday鈥檚 constant (96485 C/mol).

The term \(Q\) is called the reaction quotient, represented by:
  • \(Q = \frac{\text{[products]}}{\text{[reactants]}}\)
For cells under non-standard conditions, the Nernst Equation lets us see how changes in the concentrations of reactants and products affect the cell potential. In real-world scenarios, conditions often deviate from standard, making this equation invaluable for accurate predictions.
Standard Electrode Potential
Standard electrode potential \(E^\circ\) is a measure of the individual potential of a reversible electrode at standard state, which is 298 K, 1 M concentration for solutions, and a pressure of 1 atm. It provides insights into the tendency of a half-cell to be reduced; higher values indicate a greater tendency to gain electrons, or be reduced.

Electrochemical cells are composed of two half-cells. For instance, in the exercise, the zinc electrode involves oxidation:
  • \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2\text{e}^- \)
The copper electrode involves reduction:
  • \( \text{Cu}^{2+}(aq) + 2\text{e}^- \rightarrow \text{Cu(s)} \)
The standard electrode potential of these half-reactions can be found in electrochemical series tables where \(E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.34 \, V\) and \(E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = -0.76 \, V\).

To find the standard cell potential \(E^\circ_{\text{cell}}\), use the difference between these potentials:
  • \(E^\circ_{\text{cell}} = E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} - E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = 1.10 \, V\)
This value tells us how much voltage the cell can produce under standard conditions, a vital factor for understanding the efficiency and feasibility of electrochemical cells.
Faraday's Laws
Faraday's Laws of Electrolysis are principles that quantify the amount of substance that will be deposited or dissolved at an electrode during electrolysis. These laws link the physical phenomena to the electric current flowing through the circuit.

According to Faraday's First Law, the amount of chemical change is proportional to the amount of electricity passed. The Second Law states that the amounts of different substances liberated by the same quantity of electricity passing through the electrolyte are proportional to their equivalent weights.

In our electrochemical context, we used the charge \(Q\) calculated by:
  • \(Q = I \times t\)
where \(I\) is the current and \(t\) is the time. This charge allows us to calculate the moles of electrons (using Faraday's constant) and, subsequently, determine how much of a substance is deposited.

For example, in the exercise, we calculate:
  • \(\text{Moles of electrons} = \frac{Q}{F} = \frac{225}{96485} \approx 0.00233 \, mol\)
This value tells us how many moles of \(\mathrm{Cu}^{2+}\) ions are reduced at the cathode, affecting the concentration of the solution over time. These laws provide a quantitative understanding of the changes occurring during electrolysis, critical for designing and optimizing electrochemical processes.

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Most popular questions from this chapter

Balance the following skeleton equations. The reactions occur in acidic or basic aqueous solution, as indicated. a. \(\mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{~S} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{S}_{8}\) (acidic) b. \(\mathrm{Zn}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{N}_{2} \mathrm{O} \quad\) (acidic) c. \(\mathrm{MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{4}^{-}+\mathrm{Mn} \mathrm{O}_{2} \quad\) (basic) d. \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-}+\mathrm{BrO}_{3}^{-}\) (basic)

What are the half-reactions in the electrolysis of (a) \(\mathrm{MgBr}_{2}(l) ;\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}(l)\) ?

A potassium chloride solution is electrolyzed by passing a current through the solution using inert electrodes. A gas evolves at each electrode, and there is a large increase in \(\mathrm{pH}\) of the solution. Write the half-reactions that occur at the anode and at the cathode.

A voltaic cell has an iron rod in \(0.30 M\) iron(III) chloride solution for the cathode and a zinc rod in \(0.20 M\) zinc sulfate solution for the anode. The half-cells are connected by a salt bridge. Write the notation for this cell.

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 halfreactions are $$ \begin{gathered} \mathrm{Mg}(s) \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{e}^{-} \\ \mathrm{Ni}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(s) \end{gathered} $$ 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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