Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 114

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.

Short Answer

Expert verified
The unknown metal M used to construct the standard galvanic cell is Copper (Cu). This is determined by calculating the standard electrode potential of the M²⁺/M half-cell using the given ΔG° value (-411 kJ) and the relationship \(\Delta G^\circ = -nFE^\circ\), where n is the number of moles of electrons transferred and F is the Faraday constant. The calculated standard electrode potential of 1.125 V is closest to the value for the Cu²⁺/Cu half-cell, indicating that the unknown metal is likely Copper.

Step by step solution

01

Calculate the number of moles of electrons transferred in the overall cell reaction

In the given overall cell reaction, we have 2 moles of Al³⁺ ions and 3 moles of metal ion M targeting 2 moles of Al atoms and 3 moles of M²⁺ ions. We can balance the charges on both sides to find the number of electrons transferred. The charges on the reactants side: \( 2 \times 3 = 6 \) (from 2 moles of Al³⁺) The charges on the products side: \( 3 \times 2 = 6 \) (from 3 moles of M²⁺) Both sides have a total charge of 6, so 6 moles of electrons are transferred in the overall cell reaction.
02

Use the relationship between ΔG° and the standard electrode potential

To find the standard electrode potential, E°, of the M²⁺/M half-cell, we can use the following relationship that connects ΔG° with E° and the number of moles of electrons transferred (n) in the overall cell reaction: \( \Delta G^\circ = -nFE^\circ \) where F is the Faraday constant (approximately 96485 C/mol of electrons). Using the given ΔG° value of -411 kJ, convert it to J (since F is in C/mol): ΔG° = -411 kJ × 1000 J/kJ = -411,000 J Now, we can plug in the number of moles of electrons (n) and the Faraday constant (F) to find the standard electrode potential (E°): \( E^\circ = \frac{-\Delta G^\circ}{nF} \)
03

Calculate the standard electrode potential of the M²⁺/M half-cell

Insert the values we found for ΔG°, n, and F into the equation above: \( E^\circ = \frac{-(-411,000 \text{ J})}{6 \times 96485 \text{ C/mol}} \) \( E^\circ = \frac{411,000 \text{ J}}{6 \times 96485 \text{ C/mol}} = 1.125 \text{ V} \) Thus, the standard electrode potential of the M²⁺/M half-cell is 1.125 V.
04

Identify the unknown metal

Now that we have the standard electrode potential of the M²⁺/M half-cell, we can compare it to the known redox potentials of various metals to identify M. Comparing our calculated value of 1.125 V to the standard electrode potentials of metals, we find that it is closest to the value for the Cu²⁺/Cu half-cell, which has a standard electrode potential of 1.118 V. So, it is very likely that the unknown metal M used to construct the standard galvanic cell is Copper (Cu).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

In the electrolysis of an aqueous solution of $\mathrm{Na}_{2} \mathrm{SO}_{4},$ what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{ll} {\text{}} & \quad{ \mathscr{E}^{\circ} } \\ \hline {\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}} & {2.01 \mathrm{V}} \\ {\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}} & {1.23 \mathrm{V}} \\ {2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}} & {-0.83 \mathrm{V}} \\\ {\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}} & {-2.71 \mathrm{V}}\end{array}$$

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$\mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q)$$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+}\).)

A galvanic cell consists of a standard hydrogen electrode and a copper electrode immersed in a Cu(NO \(_{3} )_{2}(a q)\) solution. If you wish to construct a calibration curve to show how the cell potential varies with \(\left[\mathrm{Cu}^{2+}\right],\) what should you plot to obtain a straight line? What will be the slope of this line?

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.

Estimate \(\mathscr{E}^{\circ}\) for the half-reaction $$2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ} :\) $$\quad\quad\quad \mathrm{H}_{2} \mathrm{O}(l)=-237 \mathrm{kJ} / \mathrm{mol}$$ $$\mathrm{H}_{2}(g)=0.0$$ $$\quad\quad\quad \mathrm{OH}^{-}(a q)=-157 \mathrm{kJ} / \mathrm{mol}$$ $$\quad \mathrm{e}^{-}=0.0$$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 18.1
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation
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.

Sign-up for free