/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 A galvanic cell is based on the ... [FREE SOLUTION] | 91影视

91影视

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 螖G and 螖G掳 is as follows: a. 螖G = 螖G掳 (because Q = 1) b. 螖G < 螖G掳 (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鈦 + e鈦 鈫 Ag) 2. H鈧侽鈧 + 2H鈦 + 2e鈦 鈫 2H鈧侽 Now, since the number of electrons on both sides is equal, we can add the half-reactions: 2Ag鈦 + H鈧侽鈧 + 2H鈦 鈫 2Ag + 2H鈧侽 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 螖G and 螖G掳
03

Case a: Concentrations Ag鈦 = 1.0 M, H鈧侽鈧 = 2.0 M, H鈦 = 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 螖G = 螖G掳. Therefore, in case a, 螖G is equal to 螖G掳.
04

Case b: Concentrations Ag鈦 = 2.0 M, H鈧侽鈧 = 1.0 M, H鈦 = 1.0 x 10鈦烩伔 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 脳 10鈦宦光伌) = 1.0 脳 10鹿鈦 Since Q is greater than 1, we know 螖G < 螖G掳. Therefore, in case b, 螖G is smaller than 螖G掳.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Galvanic cells
Galvanic cells, also known as voltaic cells, are electrochemical cells that convert chemical energy into electrical energy through spontaneous redox reactions. They consist of two different electrodes immersed in electrolyte solutions, which are connected by a salt bridge or porous disk that allows ions to move between the solutions. This flow of ions maintains electrical neutrality.

The galvanic cell operates by separating the oxidation and reduction reactions into two half-cells, each containing a respective electrode and solution. The anode is where oxidation occurs, losing electrons, and the cathode is where reduction takes place, gaining electrons. Electrons flow from the anode to the cathode through an external circuit, generating an electric current.

In the problem, the half-reactions involve the reduction of silver ions to silver (Ag) at one electrode, and the reduction of hydrogen peroxide to water at another. By understanding the currents in these reactions, we can determine the direction and magnitude of the potential generated by the cell.
Reaction quotient
In electrochemistry, the reaction quotient, represented by the symbol \(Q\), is a measure of the relative concentrations of products and reactants at a given moment in time. It is similar to the equilibrium constant \(K\), but \(Q\) can be calculated for any set of conditions, not just at equilibrium.

For a given reaction, the expression for \(Q\) involves the concentration of the products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their coefficients:
  • \(Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)
In the exercise, for the reaction:
  • \(2\mathrm{Ag}^+ + \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ \rightarrow 2\mathrm{Ag} + 2\mathrm{H}_2\mathrm{O}\)
The reaction quotient \(Q\) is given by:
  • \(Q = \frac{[\mathrm{Ag}]^2 [\mathrm{H}_2\mathrm{O}]^2}{[\mathrm{Ag}^+]^2 [\mathrm{H}_2\mathrm{O}_2] [\mathrm{H}^+]^2}\)
Understanding \(Q\) helps predict the direction of the reaction. If \(Q < K\), the forward reaction is favored. If \(Q > K\), the reverse reaction occurs. And if \(Q = K\), the system is at equilibrium.
Thermodynamics
Thermodynamics in the context of electrochemistry is about the energy changes that occur during redox reactions. A critical part of this is the Gibbs free energy change \(\Delta G\), which indicates whether a reaction is spontaneous. In a galvanic cell, \(\Delta G\) is related to the cell potential (voltage) \(E\) by the equation:
  • \(\Delta G = -nFE\)
where \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday鈥檚 constant.

The standard free energy change \(\Delta G^\circ\) applies under standard conditions (1 M concentration, 1 atm pressure, 25掳C). However, reactions often occur at non-standard conditions, so we use the relationship:
  • \(\Delta G = \Delta G^\circ + RT\ln(Q)\)
where \(R\) is the gas constant and \(T\) is the temperature in Kelvin.

By calculating \(Q\) and determining how \(\Delta G\) compares to \(\Delta G^\circ\), we can predict if a cell reaction will proceed and potentially generate energy. In the exercise, different conditions show how \(Q\) affects \(\Delta G\), helping us understand the behavior of the cells.
Half-reactions
Half-reactions are a way to separate the oxidation and reduction processes of a redox reaction. Each half-reaction shows either the loss of electrons (oxidation) or the gain of electrons (reduction). In a galvanic cell, these two half-reactions take place in separate compartments, or half-cells.

The purpose of writing and balancing half-reactions is to show clearly how electrons are transferred between species. This is essential for understanding the flow of electricity in an electrochemical cell. For the given exercise:
  • The reduction half-reaction for silver ions is: \(\mathrm{Ag}^+ + \mathrm{e}^- \rightarrow \mathrm{Ag}\)
  • The reduction half-reaction for hydrogen peroxide is: \(\mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ + 2\mathrm{e}^- \rightarrow 2\mathrm{H}_2\mathrm{O}\)
Once both half-reactions are balanced for charge and mass, they can be combined to give the overall cell reaction. This balanced reaction helps us calculate cell potentials and evaluate how the cell performs under different conditions, crucial for determining its efficiency.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The free energy change for a reaction, \(\Delta G\), is an extensive property. What is an extensive property? Surprisingly, one can calculate \(\Delta G\) from the cell potential, \(\mathscr{b}\), for the reaction. This is surprising because \(\mathscr{B}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

Consider the electrolysis of a molten salt of some metal. What information must you know to calculate the mass of metal plated out in the electrolytic cell?

Although aluminum is one of the most abundant elements on earth, production of pure Al proved very difficult until the late 1800 s. At this time, the Hall- Heroult process made it relatively easy to produce pure Al. Why was pure Al so difficult to produce and what was the key discovery behind the Hall-Heroult process?

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 115\()\) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{Cr} \mathrm{O}_{4}\). The \(\mathscr{C}^{\circ}\) value for the following halfreaction 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 \(\mathscr{E}_{\text {cell }}\) 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 \(\left[\mathrm{Cr} \mathrm{O}_{4}{ }^{2-}\right]\). 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 \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\).

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

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.