Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 65

A voltaic cell is constructed that uses the following reaction and operates at \(298 \mathrm{K} :\) $$ \mathrm{Zn}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Ni}(s) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ni}^{2+}\right]=3.00 M\) and \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?(\mathbf{c})\) What is the emf of the cell when \(\left[\mathrm{Ni}^{2+}\right]=0.200 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)

Short Answer

Expert verified
(a) The emf of the cell under standard conditions is 0.53 V. (b) The emf of the cell when \([Ni^{2+}] = 3.00 \, M \) and \([Zn^{2+}] = 0.100 \, M \) is 0.56 V. (c) The emf of the cell when \([Ni^{2+}] = 0.200 \, M \) and \([Zn^{2+}] = 0.900 \, M \) is 0.42 V.

Step by step solution

01

Look up standard reduction potentials

Check the standard reduction potentials of Zn²⁺ and Ni²⁺. They are: Zn²⁺/Zn: -0.76 V Ni²⁺/Ni: -0.23 V Step 2: Calculate the standard cell potential
02

Find the value of the cell potential at standard conditions.

The overall cell potential at standard condition is given by $$E°_{cell} = E°_{cathode} - E°_{anode} $$ $$E°_{cell} = (-0.23 \, \text{V}) - (-0.76 \, \text{V}) = 0.53 \, \text{V}$$ (a) The emf of the cell under standard conditions is 0.53 V. Step 3: Apply the Nernst Equation for given concentrations
03

Calculate the emf for given concentrations using the Nernst equation.

For part (b), we are given concentrations of \([\mathrm{Ni}^{2+}] = 3.00 \, M \) and \([\mathrm{Zn}^{2+}] = 0.100 \, M \). The Nernst equation is used for this calculation: $$E = E° - \dfrac{RT}{nF}\ln Q$$ First, calculate the reaction quotient Q for the given concentrations: $$Q = \dfrac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]} = \dfrac{0.100}{3.00}$$ Now, substitute the values into the Nernst equation, with n = 2 for the exchange of 2 electrons in the reaction. $$E = 0.53 - \dfrac{(8.314)(298)}{(2)(96485)}\ln \left(\dfrac{0.100}{3.00}\right)$$ $$E = 0.56 \, \text{V}$$ (b) The emf of the cell when \([Ni^{2+}] = 3.00 \, M \) and \([Zn^{2+}] = 0.100 \, M \) is 0.56 V. For part (c), we are given concentrations of \([Ni^{2+}] = 0.200 \, M \) and \([Zn^{2+}] = 0.900 \, M \). Reapply the Nernst equation with the new given concentrations. First, calculate the reaction quotient Q for the given concentrations: $$Q = \dfrac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]} = \dfrac{0.900}{0.200}$$ Next, substitute the values into the Nernst equation: $$E = 0.53 - \dfrac{(8.314)(298)}{(2)(96485)}\ln \left(\dfrac{0.900}{0.200}\right)$$ $$E = 0.42 \, \text{V}$$ (c) The emf of the cell when \([Ni^{2+}] = 0.200 \, M \) and \([Zn^{2+}] = 0.900 \, M \) is 0.42 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.

Start your free trial

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

Key Concepts

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

Understanding Standard Reduction Potentials
Grasping the concept of standard reduction potentials is crucial when we delve into the workings of electrochemical cells like the voltaic cell in our exercise. These potentials, measured in volts, indicate the tendency of a chemical species to acquire electrons and thereby be reduced. Each half-cell reaction comes with its own standard reduction potential value, which has been determined under standard conditions: a pressure of 1 atm, temperature of 25°C (298K), and 1M concentration for all solutions.

The standard reduction potentials for zinc and nickel ions are essential for determining the cell potential. Positive values suggest a greater tendency for reduction, while negative values indicate less inclination. Recall that we had the values for Zn²⁺ and Ni²⁺ which show that Ni²⁺ has a higher tendency to gain electrons compared to Zn²⁺. Understanding these principles allows us to predict which direction electrons will flow in the cell and ultimately calculate the voltage that the cell can produce.
Determining Cell Potential
The cell potential, also known as electromotive force (EMF), is what drives the electric current in a voltaic cell. It's a measure of the energy per unit charge that is available from the redox reaction occurring within the cell. For our voltaic cell example, the EMF under standard conditions can be calculated using the standard reduction potentials of the half-reactions.

We apply the formula: \(E°_{cell} = E°_{cathode} - E°_{anode}\). Here, the cathode is the electrode where reduction takes place and the anode is where oxidation occurs. Positive cell potential means the reaction is spontaneous, which is the case with our voltaic cell, resulting in a 0.53 V under standard conditions. Handling these calculations is key to understanding the potential energy changes during redox reactions.
The Nernst Equation and Its Applications
Moving beyond standard conditions, the Nernst equation allows us to estimate the cell potential for non-standard conditions—where temperatures, pressures, or concentrations differ from the standard values. This equation integrates the standard cell potential with the reaction quotient (Q), temperature (T), the number of moles of electrons transferred in the reaction (n), the gas constant (R), and the Faraday constant (F) to calculate the actual cell potential.

In simple terms, the Nernst equation tells us how sensitive a cell's EMF is to changes in substance concentrations. For instance, changing the concentrations of either Zn²⁺ or Ni²⁺ ions will shift the cell potential. In our exercise, we used this equation to find that varying concentrations of Ni²⁺ and Zn²⁺ ions resulted in different EMFs, clearly demonstrating the direct influence concentrations have on the cell's potential.
Interpreting the Reaction Quotient (Q)
The reaction quotient (Q) is calculated like the equilibrium constant (K), but for a reaction that has not necessarily reached equilibrium. Q represents the ratio of product activities to reactant activities at any point in time during a reaction. For the voltaic cell we analyzed, Q is the ratio of the concentrations of Zn²⁺ ions to Ni²⁺ ions.

When we introduce the reaction quotient into the Nernst equation, we are accounting for how the reaction's position (relative to equilibrium) affects the cell potential. The larger the product of the reaction quotient, the smaller the cell's EMF becomes—we saw this when the EMF decreased as the concentration of Zn²⁺ increased relative to Ni²⁺. Mastery of the reaction quotient is fundamental, as it unveils the dynamic nature of cell potential based on the shifting landscapes of reactants and products.

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) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{K},\) and calculate the equilibrium constant \(K\) at 298 \(\mathrm{K}\) (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q)\) . (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(l) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q) .\)

A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{3},\) and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\) . The overall cell reaction is $$ \mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s) $$ (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 silver electrode to the iron electrode or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{K} :\) $$ \begin{array}{l}{\text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s)} \\ {\text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g)} \\ {\text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \rightarrow} \\ {\quad 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l)}\end{array} $$

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

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