/*! 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 166 Use the following data at \(25^{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the following data at \(25^{\circ} \mathrm{C}\) for the questions given below \(\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}(\mathrm{s}) \quad \mathrm{E}^{\circ}=-0.28 \mathrm{~V}\) \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Mg}(\mathrm{s}) \quad \mathrm{E}^{\circ}=-2.37 \mathrm{~V}\) What is \(\mathrm{K}\) for the equilibrium at \(25^{\circ} \mathrm{C}\) ? a. \(4 \times 10^{-70}\) b. \(4 \times 10^{70}\) c. \(2 \times 10^{-35}\) d. \(2 \times 10^{70}\)

Short Answer

Expert verified
The equilibrium constant \(K\) is \(2 \times 10^{70}\), option d.

Step by step solution

01

Write the overall cell reaction

To find the equilibrium constant (\(K\)), first determine the overall reaction by combining the given half-reactions. Reverse the more negative electrode reaction (more negative \(E^\circ\)) to serve as the oxidation process: - For \(\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}\), \(E^\circ = -0.28\text{ V}\) (cathode, reduction)- Reverse \( \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg}\), \(E^\circ = -2.37\text{ V}\), to get the oxidation half-reaction: \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \).
02

Calculate the standard cell potential (E°cell)

Next, we calculate \(E^\circ_{\text{cell}}\). Use the equation: \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\).- Cathode (Ni): \(-0.28 \text{ V}\)- Anode (Mg, reversed): \(-2.37 \text{ V}\)\[E^\circ_{\text{cell}} = -0.28 - (-2.37) = 2.09\text{ V}\].
03

Use the Nernst equation to find the equilibrium constant (K)

Use the relation between \(E^\circ_{\text{cell}}\) and \(K\): \[E^\circ_{\text{cell}} = \frac{RT}{nF}\ln K\]Given \(T = 298 \text{ K}\) and assuming \(R = 8.314 \text{ J/mol K}\), \(n = 2\), \(F = 96485 \text{ C/mol}\).\(E^\circ_{\text{cell}} = 2.09\text{ V}\)\(R\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}\)Substitute: \[2.09 = \frac{8.314 \times 298}{2 \times 96485}\ln K\]Solve for \(\ln K\): \[\ln K = \frac{2 \times 96485 \times 2.09}{8.314 \times 298}\]\[\ln K \approx 161.37\]Exponentiate: \[K \approx e^{161.37} \approx 2 \times 10^{70}\].
04

Compare calculated K with options

From the options provided, compare the calculated \(K\) to:a. \(4 \times 10^{-70}\)b. \(4 \times 10^{70}\)c. \(2 \times 10^{-35}\)d. \(2 \times 10^{70}\).The computed \(K\) value of \(2 \times 10^{70}\) matches option d.

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 91Ó°ÊÓ!

Key Concepts

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

Nernst Equation
The Nernst Equation is an essential tool in electrochemistry, connecting the cell potential of an electrochemical cell to the concentrations of the reactants and products involved in the redox reaction. This equation allows for the calculation of cell potentials under non-standard conditions. The equation is given by: \[ E = E^\circ - \frac{RT}{nF} \ln Q \]where:
  • E is the cell potential at non-standard conditions.
  • E\circ is the standard cell potential.
  • R is the universal gas constant (8.314 J/mol K).
  • T is the temperature in Kelvin.
  • n is the number of moles of electrons transferred in the reaction.
  • F is Faraday's constant (96485 C/mol), representing the charge of one mole of electrons.
  • Q is the reaction quotient, representing the ratio of the concentrations of the products to the reactants.
At equilibrium, the cell potential E is zero because no work can be done, and thus \[ E^\circ = \frac{RT}{nF} \ln K \]This form of the equation is used to calculate the equilibrium constant \( K \), as shown in the original exercise.
Standard Cell Potential
The standard cell potential, denoted as \(E^\circ_{\text{cell}}\), is the difference in potential energy between two electrodes when the concentrations of all the components are at their standard states (usually 1 M concentration for aqueous solutions and gases at 1 atm). In the given exercise, the standard cell potential is calculated from the reduction and oxidation half-reactions of magnesium and nickel.In electrochemical cells:
  • The cathode is where reduction occurs, and its potential should be negative in spontaneous galvanic cells.
  • The anode is where oxidation takes place, typically having a more negative potential.
For the exercise problem, the standard potential of the nickel electrode reaction, \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \), is \(-0.28 \text{ V}\), and is used as the cathode potential. The magnesium has a more negative standard potential at \(-2.37 \text{ V}\), and after reversing its reaction becomes the anode:\[ \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \]The standard cell potential is calculated by subtracting the anode potential from the cathode's:\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.28 - (-2.37) = 2.09 \text{ V} \]
Half-Reactions in Electrochemistry
Half-reactions are an integral part of understanding electrochemical cells, as they break down a redox reaction into its oxidation and reduction components. These half-reactions detail which species are gaining electrons and which are losing them.In the original exercise, the half-reactions are critical for finding the overall cell reaction:
  • The reduction half-reaction: \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \) occurs at the cathode.
  • The oxidation half-reaction: \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \) occurs at the anode.
They help in calculating the net potential of an electrochemical cell by providing the necessary electrode potentials for each half of the cell's operation. Recognizing and balancing these half-reactions efficiently is crucial for analyzing and predicting the behavior of electrochemical cells.
Redox Reactions
Redox reactions, or reduction-oxidation reactions, involve the transfer of electrons between two substances—one that gets reduced and one that gets oxidized. These reactions are pivotal in electrochemistry, forming the basis for energy conversion in batteries and other electrochemical devices.In every redox reaction:
  • Reduction refers to the gain of electrons, leading to a decrease in oxidation state.
  • Oxidation refers to the loss of electrons, resulting in an increase in oxidation state.
For example, in the \(\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}\) reaction, nickel ions gain electrons and undergo reduction, while\(\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-\) involves magnesium losing electrons and being oxidized. Identifying which species are reduced or oxidized helps determine the flow of electrons and the cell reactions in electrochemical contexts. The ability to balance these reactions is a critical skill in predicting the feasibility and potential of electrochemical cells.

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 standard reduction potentials of \(\mathrm{Zn}\) and Ag in water are \(\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \leftrightarrow \mathrm{Zn}\); \(\left(\mathrm{E}^{\circ}=-0.76 \mathrm{~V}\right)\) and \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \leftrightarrow \mathrm{Ag}\) \(\left(\mathrm{E}^{\circ}=+0.80 \mathrm{~V}\right)\) at \(298 \mathrm{~K}\). Which of the following reaction is/are not feasible? a. \(\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}(\mathrm{s})+\mathrm{Ag}(\mathrm{s})\) b. \(\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) \rightarrow 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})\) c. \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq})\) d. \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}\) (s)

The cell reaction for a lead storage battery is $$ \begin{array}{r} \mathrm{Pb}(\mathrm{s})+\mathrm{PbO}_{2}(\mathrm{~s})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{HSO}_{4}^{-}(\mathrm{aq}) \\ \rightarrow 2 \mathrm{~Pb}\left(\mathrm{SO}_{4}\right)(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \mathrm{E}^{\circ}=+1.92 \mathrm{~V} \end{array} $$ To provide a potential of about \(12 \mathrm{~V}\), one could a. Connect six cells in series. b. Adjust the \(\mathrm{pH}\) to 12 . c. Adjust the \(\mathrm{pH}\) to 1 . d. Greatly increase the surface area of the \(\mathrm{Pb}\) (s) and \(\mathrm{PbO}_{2}(\mathrm{~s})\).

In the following question two statements (Assertion) A and Reason (R) are given. Mark a. if \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. if \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of A; c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. A is false but \(\mathrm{R}\) is true, e. A and \(R\) both are false. (A): When acidified zinc sulphate solution is electrolyzed between zinc electrodes. It is zinc that is deposited at the cathode and hydrogen evolution does not take place (R): The electrode potential of zinc is more negative than hydrogen as the overvoltage for hydrogen evolution on zinc is quite large.

How many moles of electrons, are transferred in the following reduction- oxidation reaction? \(2 \mathrm{MnO}_{4}^{-}(\mathrm{aq})+16 \mathrm{H}^{+}(\mathrm{aq})+10 \mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{Mn}^{2+}(\mathrm{aq})+5 \mathrm{Cl}_{2}(\mathrm{~g})+8 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ? $$ a. 2 b. 5 c. 10 d. 12

Which of the following statements is/are not true for the electrochemical Daniel cell? a. Current flows from zinc electrode to copper electrode b. Electrons flow from copper electrode to zinc electrode c. Cations move toward zinc electrode d. Cations move toward copper electrode

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.