/*! 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 96 Predict whether the following re... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\mathrm{Sn}^{4+}\) ).

Short Answer

Expert verified
After identifying the half-reactions, finding the standard reduction potentials (E°) for the half-reactions, and calculating the standard cell potentials (E°) for each reaction, we can predict the spontaneity of the reactions as follows: (a) The oxidation of Sn to \(\mathrm{Sn}^{2+}\) by \(\mathrm{I}_{2}\) to form \(\mathrm{I}^{-}\): Non-spontaneous (b) The reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{I}^{-}\) to form \(\mathrm{I}_{2}\): Spontaneous (c) The reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\): Spontaneous (d) The reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) by \(\mathrm{Sn}^{2+}\) to form \(\mathrm{Sn}^{4+}\): Spontaneous

Step by step solution

01

Identify half-reactions

First, we need to identify the half-reactions involved in each of the given reactions. In the given reactions, we have: (a) oxidation of Sn to \(\mathrm{Sn}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\left.\mathrm{I}^{-}\right)\): - \(\mathrm{Sn} \to \mathrm{Sn}^{2+} + 2 e^-\) - \(2 \mathrm{I}^{-} + 2 e^- \to \mathrm{I}_{2}\) (b) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{I}^{-}\left(\right.\) to form \(\left.\mathrm{I}_{2}\right)\): - \(\mathrm{Ni}^{2+} + 2 e^- \to \mathrm{Ni}\) - \(2 \mathrm{I}^{-} \to \mathrm{I}_{2} + 2 e^-\) (c) reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\): - \(\mathrm{Ce}^{4+} + e^- \to \mathrm{Ce}^{3+}\) - \(\mathrm{H}_{2} \mathrm{O}_{2} + 2 e^- \to 2 \mathrm{OH}^{-}\) (d) reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) by \(\mathrm{Sn}^{2+}\) (to form \(\mathrm{Sn}^{4+}\)): - \(\mathrm{Cu}^{2+} + 2 e^- \to \mathrm{Cu}\) - \(\mathrm{Sn}^{2+} \to \mathrm{Sn}^{4+} + 2 e^-\)
02

Find standard reduction potentials

Next, we need to find the standard reduction potentials (E°) for the half-reactions involved in each reaction. We can either use a textbook or an online resource to find the values for each half-reaction.
03

Calculate the standard cell potential for each reaction

To predict spontaneity, we need to calculate the standard cell potential (E°) for each reaction. For a given redox reaction, this can be done using the equation: E° (cell) = E° (reduction) - E° (oxidation) where E° (reduction) and E° (oxidation) are the standard reduction potentials for the reduction half-reaction and the oxidation half-reaction, respectively. Calculate the standard cell potential for each given reaction using the standard reduction potentials found in Step 2.
04

Predict spontaneity of reactions

Finally, use the calculated E° values to predict whether each reaction will be spontaneous in acidic solution under standard conditions: - If E° (cell) > 0, the reaction is spontaneous. - If E° (cell) < 0, the reaction is non-spontaneous. Predict the spontaneity of each reaction based on their calculated E° values.

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Key Concepts

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

Standard Reduction Potential
Understanding the concept of standard reduction potential is crucial when examining spontaneous redox reactions. Simply put, it is a measure used in chemistry to predict the tendency of a chemical species to acquire electrons and thereby be reduced. Each half-reaction in a redox process has an associated standard reduction potential, expressed in volts (V).

These potentials are tabulated under standard conditions, which include a temperature of 25°C, a 1 M concentration for each ion participating in the reaction, and a partial pressure of 1 atm for any gases involved.

The more positive the standard reduction potential, the greater the species' affinity for electrons – or in other words, the stronger its oxidizing power. Conversely, a more negative potential indicates a stronger reducing agent. This information provides a quantifiable way to compare the oxidizing and reducing strengths of different species involved in redox reactions.

When predicting the spontaneity of a reaction using standard reduction potentials, we compare the potential of the reducing agent to the potential of the oxidizing agent. The reaction tends to proceed spontaneously if the overall standard cell potential is positive. To find this, as indicated in Step 2 of the exercise solution, the standard reduction potential for the oxidation half-reaction is subtracted from the reduction half-reaction.
Half-Reactions in Redox Chemistry
A deeper dive into half-reactions reveals the beauty of redox chemistry. Redox reactions are essentially made up of two parts: the reduction half and the oxidation half. These half-reactions are essential elements in diagnosing the full reaction and analyzing each participant's role.

The process of reduction involves a substance gaining electrons, whereas oxidation involves losing them. It's helpful to keep in mind the mnemonic 'OIL RIG' – Oxidation Is Loss, Reduction Is Gain – to remember which is which.

By examining half-reactions, as presented in Step 1 of our example solution, students can effectively balance redox reactions and determine how electrons are transferred between species. Identifying the substances being oxidized and reduced allows us to understand the directional flow of electrons and the conversion of chemical energy.

Moreover, recognizing half-reactions serves as a foundation for predicting whether a certain reaction can occur. It also sets the stage for understanding electrochemical cells, where these half-reactions physically occur in separate compartments but are connected through the movement of electrons.
Predicting Reaction Spontaneity
Predicting whether a chemical reaction will occur spontaneously is a fundamental skill in chemistry, and one that hinges on grasping the concepts of redox reactions and standard reduction potentials. As highlighted in Step 3 and Step 4 of the provided solution, the standard cell potential (E° cell) is pivotal for this prediction. It is determined by calculating the difference between the standard reduction potentials of the participating half-reactions.

A positive E° cell value indicates that a reaction is spontaneous, meaning it can take place without added energy once initiated. However, if the E° cell value is negative, the reaction is considered non-spontaneous, requiring external energy to proceed.

It is important to note that this prediction assumes standard conditions. Reactions can behave differently in non-standard conditions due to changes in concentrations, temperature, or pressure. This versatility is of great practical significance in fields such as electrochemistry, environmental science, and energy technology. Understanding the correlation between standard reduction potentials and the spontaneity of a reaction remains a key competency for students learning about redox processes.

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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.

Metallic magnesium can be made by the electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) What mass of \(\mathrm{Mg}\) is formed by passing a current of \(4.55\) A through molten \(\mathrm{MgCl}_{2}\), for \(3.50\) days? (b) How many minutes are needed to plate out \(10.00 \mathrm{~g} \mathrm{Mg}\) from molten \(\mathrm{MgCl}_{2}\), using \(3.50 \mathrm{~A}\) of current?

Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8} \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ., Vol. \(50,1973,61\) ). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage ( \(2 \mathrm{mi}\) ) is \(300 \mathrm{~atm} ?\) (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

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 \(\left[\mathrm{Ni}^{2+}\right]=3.00 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ni}^{2+}\right]=0.200 M\), and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)

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