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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 \(\mathrm{NH}_{3} )\) . $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \qquad K=1.0 \times 10^{13}$$

Short Answer

Expert verified
The cell potential at 25°C is 0.442 V, and to achieve a cell potential of 0.52 V, 0.454 moles of NH3 must be added.

Step by step solution

01

Write the balanced equation for the cell half-reactions

We are given the two half-reactions: \[\mathrm{Ag}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Ag}(s) \qquad \mathscr{E}^{\circ}=0.80 \mathrm{V}\] \[\mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(s) \qquad \mathscr{E}^{\circ}=0.34 \mathrm{V}\]
02

Find the standard cell potential

The standard cell potential is the difference between the standard half-cell potentials: \[\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode}\] Here, the silver half-reaction is the cathode and the copper half-reaction is the anode: \[\mathscr{E}^{\circ}_{cell} = 0.80 \mathrm{V} - 0.34 \mathrm{V} = 0.46 \mathrm{V}\]
03

Use the Nernst equation to find the cell potential

The Nernst equation is given by: \[\mathscr{E}_{cell} = \mathscr{E}^{\circ}_{cell} - \frac{RT}{nF} \ln Q\] For this cell, the temperature is 25°C (298 K) and the reaction quotient, Q, involves the concentrations of silver and copper ions. The valves for R and F are 8.314 J/mol K and 96485 C/mol, respectively. We also need to find the number of moles of electrons transferred in the reaction, n. From the balanced half-reactions, we can see that the silver half-reaction transfers 1 mole of electrons, while the copper half-reaction transfers 2 moles of electrons. Hence, the total number of moles of electrons transferred in the cell reaction is two, n=2. Now we can plug in the values for Q, R, T, and F, into the Nernst equation: \[\mathscr{E}_{cell} = 0.46 \mathrm{V} - \frac{(8.314 \mathrm{J/mol\ K})(298 \mathrm{K})}{(2)(96485 \mathrm{C/mol})} \ln \left(\frac{1.6 \times 10^{-10}}{2}\right)\] Calculating this expression, we find: \[\mathscr{E}_{cell} = 0.442 \mathrm{V}\] #b. Calculate moles of NH3 required#
04

Find the concentration of Cu(NH3)42+ at the desired cell potential

We are given that we want a new cell potential of 0.52 V. Using the Nernst equation again, we can solve for the new reaction quotient, Q, at this potential: \[\mathscr{E}_{cell} = 0.52 \mathrm{V} = \mathscr{E}^{\circ}_{cell} - \frac{RT}{nF} \ln Q_{new}\] Rearranging for Q, we have: \[Q_{new} = e^{(\frac{nF(\mathscr{E}_{cell} - \mathscr{E}^{\circ}_{cell})}{RT})} = e^{(\frac{(2)(96485 \mathrm{C/mol})(0.52 \mathrm{V} - 0.46 \mathrm{V})}{(8.314 \mathrm{J/mol\ K})(298 \mathrm{K})})}\] Calculating this expression, we find: \[Q_{new} = 1.63 \times 10^{10}\]
05

Calculate moles of NH3 needed

We are given the equilibrium constant for the formation of the Cu(NH3)42+ complex: \[K = \frac{[\mathrm{Cu}(\mathrm{NH_{3})_{4}]^{2+}}{[\mathrm{Cu}^{2+}][\mathrm{NH}_{3}]^4} = 1.0 \times 10^{13}\] We assumed that the reaction occurs in 1 L of 2.0 M Cu2+. Initially, there is no Cu(NH3)42+ complex, so the initial concentration of Cu2+ is 2.0 M. We want to add enough NH3 to give a new cell potential of 0.52 V, which we found corresponds to a reaction quotient, Q_new, of 1.63 x 10^{10}. Using Q_new, we can solve for the final concentration of Cu(NH3)42+: \[[\mathrm{Cu}(\mathrm{NH_{3})_{4}]^{2+}] = \frac{Q_{new}[\mathrm{Cu}^{2+}]}{[\mathrm{NH}_{3}]^4} = \frac{(1.63 \times 10^{10})(2 \mathrm{M})}{[\mathrm{NH}_{3}]^4}\] We can then use the equilibrium constant to find the amount of NH3 needed: \[1.0 \times 10^{13} = \frac{[\mathrm{Cu}(\mathrm{NH_{3})_{4}]^{2+}}{[\mathrm{Cu}^{2+}][\mathrm{NH}_{3}]^4}\] Solving for the moles of NH3 required: \[[\mathrm{NH}_{3}] = \sqrt[4]{\frac{[\mathrm{Cu}(\mathrm{NH_{3})_{4}]^{2+}}{K[\mathrm{Cu}^{2+}]}} = \sqrt[4]{\frac{(1.63 \times 10^{10})(2 \mathrm{M})}{(1.0 \times 10^{13})(2 \mathrm{M})}}\] Calculating this expression, we find: \[[\mathrm{NH}_{3}] = 4.54 \times 10^{-1} \mathrm{M}\] Now, we can find how many moles of NH3 are needed by multiplying this concentration by the volume of the solution (1.0 L): \[Moles \, NH_3 = (4.54 \times 10^{-1} \mathrm{M})(1.0 \mathrm{L}) = 4.54 \times 10^{-1} \mathrm{mol} \, NH_3\] In conclusion, the cell potential at 25°C is 0.442 V, and 0.454 moles of NH3 must be added to achieve a cell potential of 0.52 V.

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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 gives us a way to calculate the cell potential of an electrochemical cell under non-standard conditions. It's a pivotal formula in electrochemistry, allowing us to see how the voltage of a cell changes with varying concentrations of reactants and products. The equation is written as: \[\mathscr{E}_{cell} = \mathscr{E}^{\circ}_{cell} - \frac{RT}{nF} \ln Q\] where \(\mathscr{E}_{cell}\) is the cell potential, \(\mathscr{E}^{\circ}_{cell}\) is the standard cell potential, \(R\) is the universal gas constant (8.314 J/mol K), \(T\) is the temperature in Kelvin, \(n\) represents the number of moles of electrons exchanged, \(F\) is the Faraday constant (96485 C/mol), and \(Q\) is the reaction quotient. The Nernst equation helps us understand how the cell potential can be affected by changes in concentration. For example, if the concentration of reactants increases or the products decrease, we might observe a change in the cell potential. This dynamic nature of galvanic cells is an important consideration in their practical applications such as batteries.
Cell potential
The cell potential, symbolized as \(\mathscr{E}_{cell}\), is the measure of the driving force behind the direction of the electrons' flow in a redox reaction within a galvanic cell. It represents the potential energy difference between the two electrodes and indicates how much work can be produced by the cell. In a galvanic cell, electron flow occurs from the anode to the cathode. The potential is positive when calculated for a galvanic cell, meaning the reaction occurs spontaneously, converting chemical energy into electrical energy. The cell potential under standard conditions is calculated using the equation: \[\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode}\] where \(\mathscr{E}^{\circ}_{cathode}\) and \(\mathscr{E}^{\circ}_{anode}\) are the standard electrode potentials for the cathode and anode reactions, respectively.
Standard electrode potential
The standard electrode potential, denoted as \(\mathscr{E}^{\circ}\), is a measure of the tendency of a chemical species to be reduced and is measured under standard conditions. It's a crucial value assigned to each half-reaction in electrochemistry and allows for the prediction of the direction of electron flow when two half-cells are connected. Standard conditions are defined as solutes at 1 mol/l concentration, gases at 1 atm pressure, and temperature at 25°C (298 K). Standard electrode potentials are often tabulated and used to calculate the cell potential as part of the equation for galvanic and voltaic cells. Positive standard electrode potentials indicate a greater tendency to gain electrons (reduce), while negative potentials indicate a tendency to lose electrons (oxidize). These potentials are referenced to the standard hydrogen electrode (SHE), which has been assigned a potential of 0 V. It's important to note that the standard electrode potential does not change with the concentration of species in solution.
Reaction quotient
The reaction quotient, represented as \(Q\), plays an essential role in assessing the state of a reaction at any given moment relative to its equilibrium state. It's calculated using the same formula as the equilibrium constant \(K\), but with the current concentrations or pressures of the involved species instead of their concentrations at equilibrium. For a reaction \(aA + bB \rightleftharpoons cC + dD\), the reaction quotient is expressed as: \[Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}\] In the context of electrochemistry, \(Q\) is used in the Nernst equation to determine the effect of concentration changes on the cell potential. If \(Q < K\), the reaction will proceed forward to reach equilibrium. If \(Q > K\), the reaction will proceed backward. Understanding \(Q\) allows chemists to manipulate reaction conditions to optimize the performance of galvanic cells, such as batteries, by ensuring that reactions occur efficiently in the desired direction.

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Most popular questions from this chapter

In the electrolysis of a sodium chloride solution, what volume of \(\mathrm{H}_{2}(g)\) is produced in the same time it takes to produce 257 \(\mathrm{L} \mathrm{Cl}_{2}(g),\) with both volumes measured at \(50 .^{\circ} \mathrm{C}\) and 2.50 \(\mathrm{atm}\) ?

Balance the following oxidation–reduction reactions that occur in basic solution using the half-reaction method. a. \(\mathrm{PO}_{3}^{3-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{PO}_{4}^{3-}(a q)+\mathrm{MnO}_{2}(s)\) b. \(\operatorname{Mg}(s)+\mathrm{OCl}^{-}(a q) \rightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Cl}^{-}(a q)\) c. \(\mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+(a q) \rightarrow\) $$\mathrm{HCO}_{3}(a q)+\mathrm{Ag}(s)+\mathrm{NH}_{3}(a q)$$

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\operatorname{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) \quad \mathscr{E}^{\circ}=-0.444 \mathrm{V}$$ $$\operatorname{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) \qquad \quad \mathscr{E}^{\circ}=-0.126 \mathrm{V}$$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$3 \ln ^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ b. What is \(\Delta G_{i}^{\circ}\) for \(\operatorname{In}^{+}(a q)\) if \(\Delta G_{f}^{\circ}=-97.9 \mathrm{kJ} / \mathrm{mol}\) for \(\operatorname{In}^{3+}(a q) ?\)

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}{\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)} & {\mathscr{E}^{\circ}=-0.440 \mathrm{V}} \\ {2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)} & {\mathscr{E}^{\circ}=0.000 \mathrm{V}}\end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{atm},\) and a weak acid, HA, at an initial concentration of 1.00 \(\mathrm{M}\) . If the observed cell potential is 0.333 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

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 111\()\) 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: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{E}_{\text { cell } \text { and }} \Delta G\) 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 \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\).

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