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Chapter 18: Problem 63

A galvanic cell is based on the following half-reactions at \(25^{\circ} \mathrm{C}\) : $$\begin{aligned}\mathrm{Ag}^{+}+\mathrm{e}^{-} & \longrightarrow \mathrm{Ag} \\\ \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \end{aligned}$$ Predict whether \(\mathscr{G}_{\text {cell }}\) is larger or smaller than \(\mathscr{C}_{\text {cell }}^{\circ}\) for the following cases. a. \(\left[\mathrm{Ag}^{+}\right]=1.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=2.0 \mathrm{M},\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) b. \(\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=1.0 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} M\)

Short Answer

Expert verified
In summary, for the given cases a and b, the relationship between ΔG and ΔG° is as follows: a. ΔG = ΔG° (because Q = 1) b. ΔG < ΔG° (because Q > 1)

Step by step solution

01

Balancing the overall cell reaction

First, we need to balance the given half-reactions. We notice that for the two reactions to be balanced, we need to multiply the first half-reaction by 2 so that there are equal numbers of electrons: 1. 2(Ag⁺ + e⁻ → Ag) 2. H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O Now, since the number of electrons on both sides is equal, we can add the half-reactions: 2Ag⁺ + H₂O₂ + 2H⁺ → 2Ag + 2H₂O This is the balanced overall cell reaction. Step 2: Calculate the reaction quotient, Q
02

Reaction Quotient

Using the balanced overall cell reaction, let's determine the reaction quotient (Q): Q = \(\frac{[Ag]^2[H_2O]^2}{[Ag^+]^2[H_2O_2][H^+]^2}\) Step 3: Determine the relationship between ΔG and ΔG°
03

Case a: Concentrations Ag⁺ = 1.0 M, H₂O₂ = 2.0 M, H⁺ = 2.0 M

Using the given concentrations, let's calculate the reaction quotient Q for case a: Q = \(\frac{[Ag]^2[H_2O]^2}{[Ag^+]^2[H_2O_2][H^+]^2}\) = \(\frac{[1]^2[2]^2}{[1]^2[2][2]^2}\) = 1 Since Q = 1, we can infer that ΔG = ΔG°. Therefore, in case a, ΔG is equal to ΔG°.
04

Case b: Concentrations Ag⁺ = 2.0 M, H₂O₂ = 1.0 M, H⁺ = 1.0 x 10⁻⁷ M

Using the given concentrations, let's calculate the reaction quotient Q for case b: Q = \(\frac{[Ag]^2[H_2O]^2}{[Ag^+]^2[H_2O_2][H^+]^2}\) = \(\frac{[2]^2[1]^2}{[2]^2[1][1.0 \times 10^{-7}]^2}\) Q = 1 / (1.0 × 10⁻¹⁴) = 1.0 × 10¹⁴ Since Q is greater than 1, we know ΔG < ΔG°. Therefore, in case b, ΔG is smaller than ΔG°.

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Most popular questions from this chapter

The ultimate electron acceptor in the respiration process is molecular oxygen. Electron transfer through the respiratory chain takes place through a complex series of oxidation-reduction reactions. Some of the electron transport steps use iron-containing proteins called cytochromes. All cytochromes transport electrons by converting the iron in the cytochromes from the \(+3\) to the \(+2\) oxidation state. Consider the following reduction potentials for three different cytochromes used in the transfer process of electrons to oxygen (the potentials have been corrected for \(\mathrm{pH}\) and for temperature): \(\begin{aligned} \text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{2+}\right) & \\ \mathscr{B} &=0.385 \mathrm{~V} \\ \text { cytochrome } \mathrm{b}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{b}\left(\mathrm{Fe}^{2+}\right) & \\ \mathscr{E} &=0.030 \mathrm{~V} \\ \text { cytochrome } \mathrm{c}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{c}\left(\mathrm{Fe}^{2+}\right) & \\ \mathscr{Z} &=0.254 \mathrm{~V} \end{aligned}\) In the electron transfer series, electrons are transferred from one cytochrome to another. Using this information, determine the cytochrome order necessary for spontaneous transport of electrons from one cytochrome to another, which eventually will lead to electron transfer to \(\mathrm{O}_{2}\).

It took \(150 . \mathrm{s}\) for a current of \(1.25 \mathrm{~A}\) to plate out \(0.109 \mathrm{~g}\) of \(\mathrm{a}\) metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of \(1+\).

The compound with the formula \(\mathrm{TII}_{3}\) is a black solid. Given the following standard reduction potentials, \(\begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & & 8^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & & \mathscr{C}^{\circ} &=0.55 \mathrm{~V} \end{aligned}\) would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C}:\) $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 \mathrm{M}) \| \mathrm{Cu}^{2+}(2.50 \mathrm{M})\right| \mathrm{Cu}$$ The mass of each electrode is \(200 . \mathrm{g}\). a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after \(10.0 \mathrm{~A}\) of current has flowed for \(10.0 \mathrm{~h}\). (Assume each half-cell contains \(1.00 \mathrm{~L}\) of solution.) c. Calculate the mass of each electrode after \(10.0 \mathrm{~h}\). d. How long can this battery deliver a current of \(10.0 \mathrm{~A}\) before it goes dead?

What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{NiBr}_{2}\) solution b. \(1.0 \mathrm{M} \mathrm{AlF}_{3}\) solution c. \(1.0 \mathrm{M} \mathrm{MnI}_{2}\) solution
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