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Chapter 18: Problem 71

Consider the cell described below: $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M)\right|\left|\mathrm{Ag}^{+}(1.00 M)\right| \mathrm{Ag}$$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Zn}^{2+}\right]\) to have changed by 0.20 $\mathrm{mol} / \mathrm{L}\( . (Assume \)T=25^{\circ} \mathrm{C} . )$

Short Answer

Expert verified
Combining the half-cell reactions and using the Nernst equation, the cell potential after the change in concentration of \(\mathrm{Zn^{2+}}\) by 0.20 mol/L is approximately \(1.53 V\).

Step by step solution

01

Write down the half-cell reactions

The half cell reactions of the cell are given below: Zn → Zn²⁺ + 2 e⁻ (oxidation half-cell) Ag⁺ + e⁻ → Ag (reduction half-cell)
02

Determine the overall cell reaction

Add the half-cell reactions to get the overall cell reaction: Zn + 2Ag⁺ → Zn²⁺ + 2Ag
03

Calculate the initial cell potential

We use the standard electrode potentials to determine the initial cell potential. The standard electrode potentials for both half-cells are: E₀(Zn / Zn²⁺) = -0.76 V E₀(Ag⁺ / Ag) = 0.80 V The initial cell potential, E₀_cell, is determined by subtracting the oxidation potential from the reduction potential: E₀_cell = E₀(Ag⁺ / Ag) - E₀(Zn / Zn²⁺) = 0.80 V - (-0.76V) = \(1.56 V\)
04

Apply the Nernst equation

The Nernst equation allows us to calculate the cell potential when the concentrations of species involved in the reaction are not at their standard states: E_cell = E₀_cell - \(\frac{RT}{nF}\) ln(Q) Where E_cell is the cell potential, R is the gas constant \(8.314 J/(mol \times K)\), T is the temperature in Kelvin, n is the number of moles of electrons transferred in the cell reaction, F is the Faraday constant \(96485 C/mol\), and Q is the reaction quotient. Given the information in the problem, the reaction quotient Q is: Q = \(\frac{[Zn^{2+}]}{[Ag^{+}]^2}\)
05

Calculate the change in concentration

The problem states that the concentration of Zn²⁺ ions has changed by 0.20 mol/L. We need to find the new concentrations of Zn²⁺ and Ag⁺: New [Zn²⁺] = Initial [Zn²⁺] + 0.20 mol/L = 1.00 M + 0.20 M = 1.20 M New [Ag⁺] = Initial [Ag⁺] - 0.20 mol/L = 1.00 M - 0.20 M = 0.80 M Using these values, we calculate the reaction quotient, Q: Q = \(\frac{1.20}{0.80^2} = \frac{1.20}{0.64} = 1.875\)
06

Calculate T(K) and the cell potential

Convert the temperature from Celsius to Kelvin: T = 25°C + 273.15 K = 298.15 K Now, we can plug everything into the Nernst equation and calculate the cell potential after the change in concentration: E_cell = E₀_cell - \(\frac{8.314 J/(mol \times K) \times 298.15 K}{2 \times 96485 C/mol}\) ln(1.875) E_cell = \(1.56 V - \frac{8.314 \times 298.15}{2 \times 96485} \times \ln(1.875)\) E_cell ≈ \(1.56 V - 0.026 V \approx 1.53 V\) After the reaction has operated long enough for the [Zn²⁺] to have changed by 0.20 mol/L, the cell potential is approximately \(1.53 V\).

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Most popular questions from this chapter

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at $25^{\circ} \mathrm{C} :$ $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO \(_{2}\) mixture with only 0.20\(\% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

A galvanic cell is based on the following half-reactions: $$\begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \end{aligned}$$ In this cell, the silver compartment contains a silver electrode and excess AgCl(s) \(\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right),\) and the copper compartment contains a copper electrode and $\left[\mathrm{Cu}^{2+}\right]=2.0 \mathrm{M}$ . a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) . b. Assuming 1.0 \(\mathrm{L}\) of 2.0\(M \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of 0.52 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3} )\) . $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \qquad K=1.0 \times 10^{13}$$

In 1973 the wreckage of the Civil War ironclad USS Monitor was discovered near Cape Hatteras, North Carolina. [The Monitor and the CSS Virginia (formerly the USS Merrimack) fought the first battle between iron-armored ships.] In 1987 investigations were begun to see if the ship could be salvaged. It was reported in Time (June 22, 1987) that scientists were considering adding sacrificial anodes of zinc to the rapidly corroding metal hull of the Monitor. Describe how attaching zinc to the hull would protect the Monitor from further corrosion.

Consider the cell described below: $$\text { Al }\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb}$$ Calculate the cell potential after the reaction has operated long enough for the [Al \(^{3+} ]\) to have changed by 0.60 \(\mathrm{mol} / \mathrm{L}\) . (Assume \(T=25^{\circ} \mathrm{C} .\)

Consider the following galvanic cell at \(25^{\circ} \mathrm{C} :\) $$\text { Pt }\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co}$$ The overall reaction and equilibrium constant value are $$2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) \rightleftharpoons_{2 \mathrm{Cr}^{3+}}(a q)+\mathrm{Co}(s) \quad K=2.79 \times 10^{7}$$ Calculate the cell potential, \(\mathscr{E},\) for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.
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