/*! 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 595 You are given the following Dani... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 595

You are given the following Daniell cell: $$ \mathrm{Zn}, \mathrm{Zn}^{+2}(.50 \mathrm{~m}) \| \mathrm{Cu}^{2+}(0.20 \mathrm{~m}), \mathrm{Cu} $$ The concentrations of the ions are given in parentheses. Find the \(E\) for this cell at \(25^{\circ}\) c. Assume the following standard oxidation potentials: $$ \begin{array}{ll} \mathrm{Zn} \rightarrow \mathrm{Zn}^{+2}+2 \mathrm{e}^{-} & \mathrm{E}^{\circ}=.763 \\ \mathrm{Cu} \rightarrow \mathrm{Cu}^{+2}+2 \mathrm{e}^{-} & \mathrm{E}^{\circ}=-.337 \end{array} $$

Short Answer

Expert verified
The cell potential (E) of the given Daniell cell at 25掳C is approximately -1.045 V.

Step by step solution

01

Write down the Nernst equation

The Nernst equation is given by: \[ E = E^\circ - \frac{RT}{nF} \ln Q \] where E is the cell potential, \(E^\circ\) is the standard cell potential, R is the gas constant, T is the temperature in kelvin, n is the number of electrons transferred in the redox reaction, F is the Faraday constant, and Q is the reaction quotient.
02

Calculate temperature in kelvin

Given the temperature is 25掳C, we need to convert it to kelvin before using it in the Nernst equation. Temperature in kelvin (T) = 25 + 273.15 = 298.15 K
03

Write the half-reactions for Zn and Cu

The half-reactions are given by: \[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^- \] \[ \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \]
04

Combine the half-reactions to find the overall redox reaction

By combining the Zn and Cu half-reactions, we get the overall redox reaction: \[ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} +\text{Cu} \]
05

Calculate the standard cell potential

From the given standard oxidation potentials, we can find the standard cell potential (\(E^\circ\)) by subtracting the standard oxidation potential of the anode (Zn) from that of the cathode (Cu): \[ E^\circ = E_\text{cathode}^\circ - E_\text{anode}^\circ = (-0.337) - 0.763 = -1.10 \text{ V} \]
06

Determine the reaction quotient (Q)

The reaction quotient (Q) for the given redox reaction is: \[ Q = \frac{[\text{Zn}^{2+}][\text{Cu}]}{[\text{Cu}^{2+}][\text{Zn}]} \] Since the concentrations of the solid reactants (Zn and Cu) do not change, we can omit those from the expression and get: \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \] Now plug in the values, we get: \[ Q = \frac{0.50}{0.20} = 2.50 \]
07

Calculate the cell potential (E) using the Nernst equation

Now we have all the values we need to plug into the Nernst equation. The number of electrons transferred in the redox reaction (n) is 2, the gas constant (R) is 8.314 J/mol K, and the Faraday constant (F) is 96,485 C/mol. \[ E = E^\circ - \frac{RT}{nF} \ln Q \] \[ E = -1.10 - \frac{(8.314)(298.15)}{(2)(96,485)} \ln2.50 \] \[ E = -1.10 - (-0.021)\ln2.50 \] \[ E \approx -1.045 \text{ V} \] The electromotive force (EMF) or the cell potential (E) of the given Daniell cell at 25掳C is approximately -1.045 V.

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.

Daniell Cell
The Daniell cell is a specific type of electrochemical cell. It is named after the scientist John Frederic Daniell, who invented it in 1836. This cell uses a spontaneous redox reaction between zinc and copper to generate electricity.
  • The anode is the zinc electrode, where oxidation occurs. Zinc atoms lose electrons, turning into zinc ions.
  • The cathode is the copper electrode, where reduction occurs. Copper ions gain electrons and form copper metal.
A salt bridge or a porous barrier between the two half-cells allows ions to move between them. This setup completes the circuit and ensures charge balance, allowing the cell to function properly. In this system, electrons flow from the zinc electrode through an external circuit to the copper electrode, creating an electric current. The Daniell cell is one of the most fundamental examples used in electrochemistry to illustrate basic principles.
Standard Cell Potential
The standard cell potential, also denoted as \(E^\circ\), is a measure of the voltage or electric potential difference between two half-cells. It is calculated under standard conditions, which are:
  • A temperature of 298 K (25掳C)
  • 1 M concentration for all aqueous species involved
  • 1 atm pressure for any gases involved
To find the standard cell potential, we use the standard reduction potentials from the two half-reactions.
  • The cathode's standard potential is subtracted by the anode's potential.
  • For the Daniell cell: \(E^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ \).
Standard cell potentials help predict the direction of electron flow. A positive standard cell potential indicates a reaction that can proceed spontaneously under standard conditions.
Nernst Equation
The Nernst equation allows us to calculate the cell potential under non-standard conditions. This equation incorporates variables such as temperature and ion concentration differences. The general form of the equation is:\[E = E^\circ - \frac{RT}{nF} \ln Q\]Where:
  • \(E\) is the cell potential at the given conditions.
  • \(E^\circ\) is the standard cell potential.
  • \(R\) is the universal gas constant (8.314 J/(mol路K)).
  • \(T\) is the temperature in kelvin (K).
  • \(n\) is the number of moles of electrons transferred in the reaction.
  • \(F\) is the Faraday constant (approximately 96485 C/mol).
  • \(Q\) is the reaction quotient.
The Nernst equation effectively links chemical concentrations to electric potential, allowing electrochemical cells' behavior prediction beyond standard conditions.
Redox Reactions
Redox reactions are vital in electrochemistry as they involve the transfer of electrons between two chemicals. The term "redox" stems from two processes:
  • Reduction: Gain of electrons. It occurs at the cathode in electrochemical cells.
  • Oxidation: Loss of electrons. It takes place at the anode in electrochemical cells.
In the context of a Daniell cell:
  • Oxidation at the anode: Zinc transitions from elemental zinc to zinc ions, releasing electrons.
  • Reduction at the cathode: Copper ions gain electrons to form elemental copper.
By combining the half-reactions, we get the overall balanced redox reaction. Redox reactions are crucial in batteries and power the electron flow that forms the basis of electrical circuits in many devices.

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

Determine the volume in milliliters of \(20 \mathrm{M} \mathrm{KMnO}_{4}\) required to oxidize \(25.0 \mathrm{ml}\) of \(40 \mathrm{M} \mathrm{FeSO}_{4}\), in acidic solution. Assume the reaction which occurs is the oxidation of \(\mathrm{Fe}^{2+}\) by \(\mathrm{MnO}^{-}{ }_{4}\) to give \(\mathrm{Fe}^{+3}\) and \(\mathrm{Mn}^{2+}\).

Show how Le Chatelier's principle for oxidation-reduction reactions corresponds to the Nernst equation.

For the following oxidation-reduction reaction, (a) write out the two half- reactions and balance the equation, (b) calculate \(\Delta E^{\circ}\), and (c) determine whether the reaction will proceed spontaneously as written; \(\mathrm{Fe}^{2+}+\mathrm{MnO}^{-}{ }_{4}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (1) \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \leftrightarrows \mathrm{Fe}^{2+}, \mathrm{E}^{\circ}=0.77 \mathrm{eV}\) (2) \(\mathrm{MnO}^{-}{ }_{4}+8 \mathrm{H}^{+}+6 \mathrm{e}^{-} \leftrightarrows \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.51 \mathrm{eV}\)

For 1000 seconds, a current of \(0.0965\) amp is passed through a \(50 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{NaCl}\). You have only the reduction of \(\mathrm{H}_{2} \mathrm{O}\) to \(\mathrm{H}_{2}\) at the cathode and oxidation of \(\mathrm{Cl}^{-}\) to \(\mathrm{Cl}_{2}\) at the anode. Determine the average concentration of \(\mathrm{OH}^{-}\) in the final solution.

For the following voltaic cell, write the half-reactions, designating which is oxidation and which is reduction. Write the cell reaction and calculate the voltage of the cell made from standard electrodes. The cell is \(\mathrm{Co} ; \mathrm{Co}^{+2} \mid \mathrm{Ni}^{+2} ; \mathrm{Ni}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

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.