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Chapter 18: Problem 69

The overall reaction in the lead storage battery is $\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow$ \(2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) Calculate \(\mathscr{E}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 \mathrm{M}\), that is, $\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 \mathrm{M} . \mathrm{At} 25^{\circ} \mathrm{C}, \mathscr{E}^{\circ}=2.04 \mathrm{V}$ for the lead storage battery.

Short Answer

Expert verified
Using the Nernst equation and the given reaction conditions, we can find the cell potential (EMF) of the lead-acid battery at 25°C. The number of electrons transferred in the reaction is 2, and the reaction quotient Q is \(\frac{1}{[\mathrm{H}^+]^2 [\mathrm{HSO}_4^-]^2}\) which simplifies to \(\frac{1}{(4.5)^2(4.5)^2}\). Plugging all values into the Nernst equation, we find that the cell potential (E) is approximately 2.01 V.

Step by step solution

01

Write the Nernst equation

To find the EMF of the cell, we must use the Nernst equation, which relates the cell potential, standard cell potential, temperature, the number of electrons transferred, and the concentrations of reactants and products. The Nernst equation is: \[E = E^\circ - \frac{RT}{nF} \ln{Q}\] Where: - E is the cell potential (EMF) - E° is the standard cell potential (2.04 V, given) - R is the gas constant (8.314 J/mol·K) - T is the temperature in Kelvin (25°C = 298 K) - n is the number of electrons transferred in the reaction - F is the Faraday constant (96485 C/mol) - Q is the reaction quotient for the reaction at the specific concentrations
02

Find the number of electrons transferred n

Analyzing the given reaction, determine the number of electrons transferred by balancing the charges of reactants and products: \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) 2 electrons are transferred in the overall reaction. Thus, n = 2.
03

Calculate the reaction quotient Q

For the given reaction, the reaction quotient Q can be defined as: \[Q = \frac{[\mathrm{PbSO}_4]^2 [H_2O]^2}{[\mathrm{H}^+]^2 [\mathrm{HSO}_4^-]^2}\] Since PbSO₄ and H₂O are both solids, their concentration terms are omitted from the Q calculation: \[Q = \frac{1}{[\mathrm{H}^+]^2 [\mathrm{HSO}_4^-]^2}\] We are given that [H⁺] = [HSO₄⁻] = 4.5 M. Plug in these values to calculate Q: \[Q = \frac{1}{(4.5)^2(4.5)^2}\]
04

Calculate the cell potential E

Now plug all the values into the Nernst equation: \[E = 2.04 - \frac{8.314 \times 298}{2 \times 96485} \ln{\frac{1}{(4.5)^2(4.5)^2}}\] Calculate the value of E: \[E \approx 2.01 \, V\]
05

Write the final answer

The electromotive force (EMF) of the lead-acid battery at 25°C and given concentrations is approximately 2.01 V.

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Most popular questions from this chapter

The general rule for salt bridges is that anions flow to the anode and cations flow to the cathode. Explain why this is true.

Consider the following galvanic cell: Calculate the \(K_{s p}\) value for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s) .\) Note that to obtain silver ions in the right compartment (the cathode compartment), excess solid \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) was added and some of the salt dissolved.

The black silver sulfide discoloration of silverware can be removed by heating the silver article in a sodium carbonate solution in an aluminum pan. The reaction is $$3 \mathrm{Ag}_{2} \mathrm{S}(s)+2 \mathrm{Al}(s) \rightleftharpoons 6 \mathrm{Ag}(s)+3 \mathrm{S}^{2-}(a q)+2 \mathrm{Al}^{3+}(a q)$$ a. Using data in Appendix \(4,\) calculate \(\Delta G^{\circ}, K,\) and \(\mathscr{E}^{\circ}\) for the above reaction at $25^{\circ} \mathrm{C} .\left[\text { For } \mathrm{Al}^{3+}(a q), \Delta G_{\mathrm{f}}^{\circ}=-480 . \mathrm{kJ} / \mathrm{mol.}\right]$ b. Calculate the value of the standard reduction potential for the following half-reaction: $$2 \mathrm{e}^{-}+\mathrm{Ag}_{2} \mathrm{S}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)$$

A standard galvanic cell is constructed so that the overall cell reaction is $$2 \mathrm{Al}^{3+}(a q)+3 \mathrm{M}(s) \longrightarrow 3 \mathrm{M}^{2+}(a q)+2 \mathrm{Al}(s)$$ where \(\mathrm{M}\) is an unknown metal. If \(\Delta G^{\circ}=-411 \mathrm{kJ}\) for the overall cell reaction, identify the metal used to construct the standard cell.

An electrochemical cell is set up using the following unbalanced reaction: $$\mathrm{M}^{a+}(a q)+\mathrm{N}(s) \longrightarrow \mathrm{N}^{2+}(a q)+\mathrm{M}(s)$$ The standard reduction potentials are: $$\mathrm{M}^{a+}+a \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=0.400 \mathrm{V}$$ $$\mathrm{N}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{N} \quad \mathscr{E}^{\circ}=0.240 \mathrm{V}$$ The cell contains 0.10\(M \mathrm{N}^{2+}\) and produces a voltage of 0.180 \(\mathrm{V}\) . If the concentration of \(\mathrm{M}^{a+}\) is such that the value of the reaction quotient \(Q\) is \(9.32 \times 10^{-3},\) calculate \(\left[\mathrm{M}^{a+}\right] .\) Calculate \(w_{\text { max }}\) for this electrochemical cell.
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