Write the Nernst equation for the following processes at some temperature \(T\) : (a) \(\mathrm{Mg}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+\operatorname{Sn}(s)\) (b) \(2 \mathrm{Cr}(s)+3 \mathrm{~Pb}^{2+}(a q) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{~Pb}(s)\)

Short Answer

Expert verified
The Nernst equation for the reactions given are \( E_{\mathrm{Mg}^{2+}/\mathrm{Mg}} = E_{\mathrm{Mg}^{2+}/\mathrm{Mg}}^0 - \frac{RT}{2F} ln \frac{[\mathrm{Mg}^{2+}]}{[\mathrm{Sn}^{2+}]} \) and \( E_{\mathrm{Cr}^{3+}/\mathrm{Cr}} = E_{\mathrm{Cr}^{3+}/\mathrm{Cr}}^0 - \frac{RT}{6F} ln \left( \frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Pb}^{2+}]^3} \right) \) respectively.

Step by step solution

01

Identify the Components of the Nernst equation

Recall the Nernst equation: \( E = E^0 - \frac{RT}{nF} lnQ \). Identify each component needed for this equation in each of the chemical reactions.
02

Write the Nernst Equation for the first reaction

For the reaction \( \mathrm{Mg}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+\operatorname{Sn}(s) \), the Nernst equation would be \( E_{\mathrm{Mg}^{2+}/\mathrm{Mg}} = E_{\mathrm{Mg}^{2+}/\mathrm{Mg}}^0 - \frac{RT}{2F} ln \frac{[\mathrm{Mg}^{2+}]}{[\mathrm{Sn}^{2+}]} \)
03

Write the Nernst Equation for the second reaction

For the reaction \( 2 \mathrm{Cr}(s)+3 \mathrm{~Pb}^{2+}(a q) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{~Pb}(s) \), the Nernst equation would be \( E_{\mathrm{Cr}^{3+}/\mathrm{Cr}} = E_{\mathrm{Cr}^{3+}/\mathrm{Cr}}^0 - \frac{RT}{6F} ln \left( \frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Pb}^{2+}]^3} \right) \)

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 Vaia!

Key Concepts

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

Electrochemical Reactions
Electrochemical reactions are at the heart of batteries, corrosion, and various industrial processes. They involve the movement of electrons from one species to another and occur in electrochemical cells. In simple terms, these cells have two electrodes—an anode and a cathode—immersed in an electrolyte solution. An electrochemical reaction can be broken down into two half-reactions: oxidation and reduction. Oxidation is the loss of electrons, while reduction is the gain of electrons.

The electrochemical series, a list ordered by standard electrode potentials, predicts the direction of electron flow. Species with higher potentials will tend to gain electrons and be reduced, while those with lower potentials will tend to lose electrons and be oxidized. This is essential to understanding why, in the provided example, magnesium (Mg) loses electrons to form \(\mathrm{Mg}^{2+}\) while tin (Sn) gains electrons to form metallic Sn.
Chemical Equilibrium
Chemical equilibrium is a state in which the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. It is crucial to recognize that equilibrium does not mean the reactants and products are in equal concentration, but rather that their rates of formation are balanced.

When applying the Nernst equation, equilibrium plays a critical role as the equation includes the reaction quotient \(Q\), representing the ratio of product concentrations to reactant concentrations at any given moment, not just at equilibrium. In the provided exercise, calculating the voltage of the cell requires understanding the dynamics of the reaction and how the concentrations of the ions involved (e.g., \(\mathrm{Mg}^{2+}\) and \(\mathrm{Sn}^{2+}\)) affect the cell's potential before it reaches equilibrium.
Thermodynamics in Chemistry
Thermodynamics is a branch of physics that deals with heat, work, and energy. In chemistry, thermodynamics often involves the study of energy changes during chemical reactions. The Nernst equation is a fine example where thermodynamics intersects with electrochemistry, as it relates the electrochemical potential of a cell to the concentrations of species involved in the reaction.

The term \(RT/nF\) in the Nernst equation stems from thermodynamics principles. \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred in the reaction, and \(F\) is Faraday’s constant, relating the amount of charge to moles of electrons. The equation allows us to calculate the cell potential under non-standard conditions, reflecting the second law of thermodynamics, which tells us that reactions proceed in the direction that tends to minimize free energy.

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 oxidation of \(25.0 \mathrm{~mL}\) of a solution containing \(\mathrm{Fe}^{2+}\) requires \(26.0 \mathrm{~mL}\) of \(0.0250 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in acidic solution. Balance the following equation and calculate the molar concentration of \(\mathrm{Fe}^{2+}\): $$ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+}+\mathrm{H}^{+} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+} $$

Predict whether the following reactions would occur spontaneously in aqueous solution at \(25^{\circ} \mathrm{C}\). Assume that the initial concentrations of dissolved species are all \(1.0 M\). (a) \(\mathrm{Ca}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{Cd}(s)\) (b) \(2 \mathrm{Br}^{-}(a q)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Br}_{2}(l)+\operatorname{Sn}(s)\) (c) \(2 \mathrm{Ag}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Ni}(s)\) (d) \(\mathrm{Cu}^{+}(a q)+\mathrm{Fe}^{3+}(a q) \longrightarrow$$\mathrm{Cu}^{2+}(a q)+\mathrm{Fe}^{2+}(a q)\)

How many faradays of electricity are required to produce (a) \(0.84 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at exactly \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) from aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution; (b) \(1.50 \mathrm{~L}\) of \(\mathrm{Cl}_{2}\) at \(750 \mathrm{mmHg}\) and \(20^{\circ} \mathrm{C}\) from molten \(\mathrm{NaCl}\); (c) \(6.0 \mathrm{~g}\) of Sn from molten \(\mathrm{SnCl}_{2} ?\)

A galvanic cell is constructed by immersing a piece of copper wire in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}\) solution and a zinc strip in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{ZnSO}_{4}\) solution. (a) Calculate the emf of the cell at \(25^{\circ} \mathrm{C}\) and predict what would happen if a small amount of concentrated \(\mathrm{NH}_{3}\) solution were added to (i) the \(\mathrm{CuSO}_{4}\) solution and (ii) the \(\mathrm{ZnSO}_{4}\) solution. Assume that the volume in each compartment remains constant at \(25.0 \mathrm{~mL}\). (b) In a separate experiment, \(25.0 \mathrm{~mL}\) of \(3.00 M \mathrm{NH}_{3}\) are added to the \(\mathrm{CuSO}_{4}\) solution. If the emf of the cell is \(0.68 \mathrm{~V},\) calculate the formation constant \(\left(K_{\mathrm{f}}\right)\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\).

Complete the following table. State whether the cell reaction is spontaneous, nonspontaneous, or at equilibrium. $$ \begin{array}{c|c|c} \boldsymbol{E} & \boldsymbol{\Delta} \boldsymbol{G} & \text { Cell Reaction } \\\ \hline > 0 & & \\ \hline & > 0 & \\ \hline=0 & & \\ \hline \end{array} $$

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.

Sign-up for free