Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Under what circumstances is the Nernst equation applicable? (b) What is the numerical value of the reaction quotient, \(Q\), under standard conditions? (c) What happens to the emf of a cell if the concentrations of the reactants are increased?

Short Answer

Expert verified
(a) The Nernst equation is applicable when the involved species are in equilibrium, the temperature is reasonably constant, and the reaction involves electron transfer (redox reaction). It is given by: \(E = E^\circ - \dfrac{RT}{nF} \ln Q_\mathrm{r}\). (b) Under standard conditions, the numerical value of the reaction quotient, \(Q\), is 1. (c) When the concentrations of the reactants are increased, the cell potential (emf) will also increase, indicating that the cell can provide more electrical energy with higher concentrations of reactants.

Step by step solution

01

(a) The Nernst equation and its applicability

The Nernst equation is used to determine the reduction potential of a half-cell in an electrochemical cell, taking into account the concentration or partial pressure of the involved species. It can be written as: \(E = E^\circ - \dfrac{RT}{nF} \ln Q_\mathrm{r}\) Where: - \(E\) is the cell potential (or emf) at non-standard conditions - \(E^\circ\) is the cell potential at standard conditions - \(R\) is the gas constant (\(8.314\, \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}\)) - \(T\) is the temperature in Kelvin - \(n\) is the number of moles of electrons transferred in the redox reaction - \(F\) is the Faraday's constant (\(9.6485 \times 10^4\, \mathrm{C} \mathrm{mol}^{-1}\)) - \(Q_\mathrm{r}\) is the reaction quotient The Nernst equation is applicable when: 1. The involved species are in equilibrium. 2. The temperature is reasonably constant. 3. The reaction involves electron transfer (redox reaction).
02

(b) Reaction quotient under standard conditions

Under standard conditions, the concentration of each species is 1 M, and the partial pressure of the involved gases is 1 atm. The reaction quotient Q can be defined as: \(Q = \dfrac{\prod([\text{products}]^{\text{stoichiometric coefficients}})}{\prod([\text{reactants}]^{\text{stoichiometric coefficients}})}\) Under standard conditions, the concentrations of all species are the same (1 M). Therefore, the reaction quotient Q becomes: \(Q = \dfrac{1^{\text{stoichiometric coefficients of products}}}{1^{\text{stoichiometric coefficients of reactants}}}\) Since any number raised to any power is always one, the numerical value of Q under standard conditions is: \(Q = 1\)
03

(c) Effect of increasing reactants concentrations on emf

According to the Nernst equation, the cell potential is related to the natural logarithm of the reaction quotient Q: \(E = E^\circ - \dfrac{RT}{nF} \ln Q_\mathrm{r}\) When the concentrations of the reactants are increased, the reaction quotient Q decreases (less than 1) because Q is the ratio of the concentrations of products and reactants. As a result, the value of the natural logarithm \(\ln Q_\mathrm{r}\) becomes negative. This causes the second term (\(\dfrac{RT}{nF} \ln Q_\mathrm{r}\)) in the Nernst equation to become positive. Therefore, as the concentrations of the reactants increase, the cell potential (emf) will also increase. This indicates that the cell can provide more electrical energy when the concentrations of reactants are increased.

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!

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

If you were going to apply a small potential to a steel ship resting in the water as a means of inhibiting corrosion, would you apply a negative or a positive charge? Explain.

Mercuric oxide dry-cell batteries are often used where a high energy density is required, such as in watches and cameras. The two half-cell reactions that occur in the battery are \(\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}--\rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)\) \(\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}\) (a) Write the overall cell reaction. (b) The value of \(E_{\text {red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of Sn to \(\mathrm{Sn}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\left.\mathrm{I}^{-}\right)\), (b) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{I}^{-}\left(\right.\) to form \(\left.\mathrm{I}_{2}\right),(\mathrm{c})\) reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\), (d) reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) by \(\mathrm{Sn}^{2+}\) (to form \(\mathrm{Sn}^{4+}\) ).

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q)-\cdots \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q)-\longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(\mathrm{~s})+\mathrm{OCl}^{-}(a q)-\mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{As}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{NO}_{3}^{-}(a q)-\longrightarrow\) \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)\) (acidic solution) (e) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q)-\mathrm{MnO}_{2}(s)+\mathrm{BrO}_{3}^{-}(a q)\) (basic solution) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}{ }^{2-}(a q)+\mathrm{ClO}^{-}(a q)-\cdots \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)\) (basic solution)

(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?

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