/*! 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 52 For each of the following reacti... [FREE SOLUTION] | 91影视

91影视

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 \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(I) 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 \(\mathrm{ClO}^{-}(a q)\)

Short Answer

Expert verified
\(Hg_{2}^{2+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(s) + 2Hg(l)\): E掳 = 0.384 V, 螖G掳 = -73,669 J/mol, K 鈮 \(2.94 \times 10^{129}\)

Step by step solution

01

Write a balanced equation

Write the balanced half-reactions for this process: Reduction half-reaction: \(Hg_{2}^{2+} + 2e^{-} \rightarrow 2Hg\) Oxidation half-reaction: \(2I^{-} \rightarrow I_{2} + 2e^{-}\) Now, add the two half-reactions to obtain the balanced net equation: \(Hg_{2}^{2+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(s) + 2Hg(l)\)
02

Calculate the standard emf (E掳)

Use the standard reduction potentials to calculate the standard emf for this reaction: E掳(Hg虏鈦/Hg) = 0.92 V E掳(I鈧/I鈦) = 0.536 V Using the Nernst equation, we have: E掳(cell) = E掳(reduction) - E掳(oxidation) E掳(cell) = 0.92 V - 0.536 V = 0.384 V
03

Calculate 螖G掳 at 298 K

Use the formula 螖G掳 = -nFE掳, where n is the number of moles of electrons transferred and F is Faraday's constant (96,485 C/mol): 螖G掳 = - (2)(96,485 C/mol)(0.384 V) 螖G掳 = -73,669 J/mol
04

Calculate the equilibrium constant (K) at 298 K

Use the formula 螖G掳 = -RT ln K, where R is the gas constant (8.314 J/mol K) and T is the temperature (298 K): -73,669 J/mol = -(8.314 J/mol K)(298 K) ln K ln K = 298 K 鈮 \(2.94 \times 10^{129}\) The final answers for part (a) are: Balanced equation: \(Hg_{2}^{2+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(s) + 2Hg(l)\) Standard emf (E掳): 0.384 V 螖G掳: -73,669 J/mol K: 鈮 \(2.94 \times 10^{129}\) Repeat the process for parts (b) and (c).

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 91影视!

Key Concepts

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

Electrochemistry
Electrochemistry involves the study of chemical processes that cause electrons to move. This movement of electrons is what generates electricity. In chemical reactions, these processes occur in electrochemical cells. These cells are devices that convert chemical energy into electrical energy or vice-versa. There are two main types of electrochemical cells: galvanic (or voltaic) cells, which produce electrical energy from spontaneous chemical reactions, and electrolytic cells, which use electrical energy to drive non-spontaneous reactions.

In our example, we're dealing with redox reactions where oxidation and reduction occur. Oxidation is the loss of electrons, while reduction is the gain of electrons. Each of these half-reactions takes place at electrodes. The oxidation half-reaction occurs at the anode, and the reduction half-reaction occurs at the cathode.
  • Oxidation: electrons are lost.
  • Reduction: electrons are gained.
  • Anode: where oxidation happens.
  • Cathode: where reduction happens.
Understanding these principles allows us to balance equations, calculate potentials, and produce useful energy through controlled reactions.
Standard Electrode Potential
The standard electrode potential, denoted as E掳, is the voltage, or electric potential difference, of a cell under standard conditions. Standard conditions include a temperature of 298 K, a pressure of 1 atm, and usually, solutions at 1 M concentration. These potentials allow us to predict the direction of redox reactions and calculate the standard electromotive force (emf) of a reaction.

In electrochemistry, the standard reduction potentials of different substances are used to calculate the standard emf of a cell. The emf is determined using the formula:
\[ E^{ ext{掳cell}} = E^{ ext{掳reduction}} - E^{ ext{掳oxidation}} \]
Here鈥檚 how you apply it:
  • Identify the reduction and oxidation half-reactions.
  • Consult standard tables to find the E掳 values for each half-reaction.
  • Subtract the oxidation potential from the reduction potential to find E掳 of the whole cell.
The calculated emf indicates whether a reaction will spontaneously occur. A positive emf suggests a spontaneous reaction.
Gibbs Free Energy
Gibbs Free Energy (\( \Delta G \)) helps determine whether a chemical process is spontaneous. It's the energy available to do work. In electrochemistry, \( \Delta G^{掳} \) relates to the cell potential as follows:
\[ \Delta G^{掳} = -nFE^{掳} \]

Where:
  • \( n \) is the number of moles of electrons transferred.
  • \( F \) is Faraday鈥檚 constant (96,485 C/mol).
  • \( E^{掳} \) is the standard emf in volts.
A negative \( \Delta G^{掳} \) means the reaction is spontaneous, releasing energy. For instance, if we calculate \( \Delta G^{掳} \) to be \(-73,669 \text{ J/mol}\), the reaction energetically favors products at the given conditions.

Understanding \( \Delta G\) in electrochemical reactions provides insights into reaction spontaneity and sustainability.
Equilibrium Constant
The equilibrium constant (\( K \)) conveys how far a reaction goes towards completion. In electrochemical reactions, \( K \) is related to \( \Delta G^{掳} \) through the equation:
\[ \Delta G^{掳} = -RT \ln K \]
  • \( R \) is the gas constant (8.314 J/mol K).
  • \( T \) is the temperature in Kelvin.
This equation ties the energetics of the reaction to its position at equilibrium.

Rearranging and solving for \( K \), you get:
\[ K = e^{-\Delta G^{掳}/RT} \]

A larger \( K \) value typically suggests that the products are favored at equilibrium. For example, with a calculated \( K \approx 2.94 \times 10^{129} \), the reaction proceeds nearly to completion, indicating that product formation is highly favored under standard conditions.

This understanding allows for predictions about the extent and direction of reactions.

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

Consider the half-reaction \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\) (a) Which of the lines in the following diagram indicates how the reduction potential varies with the concentration of \(\mathrm{Ag}^{+}(a q) ?(\mathbf{b})\) What is the value of \(E_{\text {red }}\) when \(\log \left[\mathrm{Ag}^{+}\right]=0 ?\)

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?

Cytochrome, a complicated molecule that we will represent as \(\mathrm{CyFe}^{2+}\), reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions (Section 19.7). At \(\mathrm{pH} 7.0\) the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}\) $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{~V} \\\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} & \longrightarrow \mathrm{CyFe}^{2+}(a q) & E_{\mathrm{red}}^{\circ} &=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air? \((\mathbf{b})\) If the synthesis of \(1.00 \mathrm{~mol}\) of ATP from adenosine diphosphate (ADP) requires a \(\Delta G\) of \(37.7 \mathrm{~kJ},\) how many moles of ATP are synthesized per mole of \(\mathrm{O}_{2} ?\)

In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(s)\) half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and \(0.050 \mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right)\). The measured cell voltage is \(1.030 \mathrm{~V}\). What is the \(\mathrm{p} K_{\mathrm{a}}\) of benzoic acid?

Is each of the following substances likely to serve as an oxidant or a reductant: \((\mathbf{a}) \mathrm{Ce}^{3+}(a q),(\mathbf{b}) \mathrm{Ca}(s),(\mathbf{c}) \mathrm{ClO}_{3}^{-}(a q),\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g) ?\)

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.