Consider a redox reaction for which \(E^{\circ}\) is a negative number. (a) What is the sign of \(\Delta G^{\circ}\) for the reaction? (b) Will the equilibrium constant for the reaction be larger or smaller than \(1 ?\) (c) Can an electrochemical cell based on this reaction accomplish work on its surroundings?

Short Answer

Expert verified
(a) Since \(E^{\circ}\) is negative, \(\Delta G^{\circ}\) will be positive for the reaction. (b) The equilibrium constant (K) for the reaction will be smaller than 1. (c) An electrochemical cell based on this reaction cannot accomplish work on its surroundings.

Step by step solution

01

(Step 1: Relate Standard Cell Potential and Gibbs Free Energy)

We will use the above relationships to express the relationship between the standard potential and Gibbs free energy: \[\Delta G^{\circ} = -nFE^{\circ}\] where \(n\) is the number of moles of electrons transferred and \(F\) is Faraday's constant.
02

(Step 2: Determine the Sign of Gibbs Free Energy)

Since \(E^{\circ}\) is given to be a negative number and both \(n\) and \(F\) are positive values, the multiplication of all three results in a positive number. Therefore, \(\Delta G^{\circ}\) is a positive number for this reaction as given: \[\Delta G^{\circ} > 0\]
03

(Step 3: Relate Gibbs Free Energy and the Equilibrium Constant)

To answer (b), we will need to express the equilibrium constant (K) in terms of Gibbs free energy. We can do this using: \[\Delta G^{\circ} = -RT\ln{K}\] where \(R\) is the ideal gas constant and \(T\) is the temperature in Kelvin.
04

(Step 4: Determine Whether the Equilibrium Constant is Greater or Smaller than 1)

Since we've determined in Step 2 that \(\Delta G^{\circ}>0\), we have: \[-RT\ln{K} > 0\] Divide both sides by \(-RT\): \[\ln{K} < 0\] Exponentiate both sides: \[K < 1\] This means that the equilibrium constant for this reaction will be smaller than 1.
05

(Step 5: Determine the Electrochemical Cell's Ability to Accomplish Work)

As the standard cell potential (\(E^{\circ}\)) is negative, this implies that the electrochemical cell is not spontaneous. Thus, an electrochemical cell based on this reaction cannot accomplish work on its surroundings. To make the reaction spontaneous, an external voltage equal to or greater than the absolute of standard potential value must be applied. In summary: (a) The sign of \(\Delta G^{\circ}\) for the reaction is positive. (b) The equilibrium constant for the reaction will be smaller than 1. (c) An electrochemical cell based on this reaction cannot accomplish work on its surroundings.

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

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) Why is an aqueous solution of \(\mathrm{MgCl}_{2}\) not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are \(96 \%\) efficient in producing the desired products in electrolysis, what mass of \(\mathrm{Mg}\) is formed by passing a current of 97,000 A for a period of 24 h?

Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons. (b) For the reaction $\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\( \)\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\( is the reducing agent and \)\mathrm{Co}^{2+}(a q)$ is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.

A voltaic cell is based on \(\mathrm{Cu}^{2+}(a q) / \mathrm{Cu}(s)\) and \(\mathrm{Br}_{2}(l) /\) \(\mathrm{Br}^{-}(a q)\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(\mathrm{C}\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\). (Thestandard entropy of \(\mathrm{Cu}^{2+}(a q)\) is $\left.S^{\circ}=-99.6 \mathrm{~J} / \mathrm{K}\right)$

A student designs an ammeter (device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for 90 min, \(32.5 \mathrm{~mL}\) of water-saturated \(\mathrm{H}_{2}(g)\) is collected. The temperature of the system is \(20^{\circ} \mathrm{C},\) and the atmospheric pressure is \(101.3 \mathrm{kPa}\). What is the magnitude of the average current in amperes?

(a) Calculate the mass of Li formed by electrolysis of molten LiCl by a current of \(7.5 \times 10^{4}\) A flowing for a period of 24 h. Assume the electrolytic cell is \(85 \%\) efficient. (b) What is the minimum voltage required to drive the reaction?

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