Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 64

A voltaic cell utilizes the following reaction: $$\mathrm{Al}(s)+3 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+3 \mathrm{Ag}(s)$$ What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution. (b) The size of the aluminum electrode is increased. (c) A solution of \(\mathrm{AgNO}_{3}\) is added to the cathode half-cell, increasing the quantity of \(\mathrm{Ag}^{+}\) but not changing its concentration. (d) \(\mathrm{HCl}\) is added to the \(\mathrm{AgNO}_{3}\) solution, precipitating some of the \(\mathrm{Ag}^{+}\) as \(\mathrm{AgCl}\).

Short Answer

Expert verified
(a) When the anode half-cell is diluted, the emf (E) of the cell will increase due to the decrease in the concentration of Al^3+ ions. (b) Increasing the size of the aluminum electrode does not affect the cell emf. (c) Adding AgNO3 solution to the cathode half-cell without changing the concentration of Ag+ ions will not affect the cell emf. (d) Precipitating some Ag+ as AgCl will decrease the cell emf as the concentration of Ag+ ions in the solution decreases.

Step by step solution

01

(a) Diluting the Anode Half-Cell

When water is added to the anode half-cell, the concentration of Al^3+ ions in the solution will decrease. According to the Nernst equation: \[ E = E^0 - \frac{0.0592}{n} \log\frac{[\mathrm{Al}^{3+}]}{[\mathrm{Ag}^+]^3} \] Decreasing the concentration of Al^3+ ions will cause the logarithmic term to decrease as well, and thus, the emf (E) will increase.
02

(b) Increasing the Size of the Aluminum Electrode

The size of the electrode does not affect the concentrations of the ionic species in the cell and therefore will not directly affect the cell emf. The emf is solely dependent on the concentrations of species involved in the redox reaction, according to the Nernst equation. So the emf of the cell remains unchanged in this situation.
03

(c) Adding AgNO3 Solution to the Cathode Half-Cell

By adding a solution of AgNO3 to the cathode half-cell, the quantity of Ag+ ions increases, but their concentration remains the same. Since the emf of the cell is determined by the concentrations of the involved species and not their absolute quantity, there will be no effect on the cell emf due to this addition.
04

(d) Precipitating Ag+ as AgCl

When HCl is added to the AgNO3 solution, it forms a precipitate with some of the Ag+ ions, producing AgCl. This process will reduce the concentration of Ag+ ions in the solution. Again, referring to the Nernst equation: \[ E = E^0 - \frac{0.0592}{n} \log\frac{[\mathrm{Al}^{3+}]}{[\mathrm{Ag}^+]^3} \] With the decrease in the concentration of Ag+ ions, the logarithmic term will increase, causing the emf (E) to decrease.

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

A voltaic cell utilizes the following reaction: $$2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q)$$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}\), \(P_{\mathrm{H}_{2}}=0.95 \mathrm{~atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M},\) and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

During the discharge of an alkaline battery, \(4.50 \mathrm{~g}\) of \(\mathrm{Zn}\) is consumed at the anode of the battery. (a) What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from \(\mathrm{Zn}\) to \(\mathrm{MnO}_{2}\) ?

(a) Write the reactions for the discharge and charge of a nickelcadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+\begin{array}{c} 2 \mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{~V} \end{array} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3},\) and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\). The overall cell reaction is $$2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s)$$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the \(\mathrm{Al}\) is not coated with its oxide.

If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(1.5 \times 10^{-4},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E_{\text {red }}^{\circ}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

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