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Given the following standard electrode potentials at \(25^{\circ} \mathrm{C}\) : \(\mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}^{2+}, \mathrm{E}^{\circ}=0.15 \mathrm{ev}\) and \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}\) \(E^{\circ}=+0.77 e \mathrm{ev}\) will the reaction \(\mathrm{Sn}^{2+}+2 \mathrm{~F} \mathrm{e}^{3+} \leftrightharpoons \mathrm{Sn}^{4+}+2 \mathrm{Fe}^{2+}\) proceed spontaneously?

Short Answer

Expert verified
The reaction \(Sn^{2+} + 2 Fe^{3+} \leftrightharpoons Sn^{4+} + 2 Fe^{2+}\) will proceed spontaneously at 25°C, as the standard electrode potential (E°) of the overall reaction is positive (+1.39 V).

Step by step solution

01

Write the half-cell reactions

Both half-cell reactions are given in the problem: 1. Sn^4+ + 2 e⁻ → Sn^2+; E° = 0.15 V 2. Fe^3+ + e⁻ → Fe^2+; E° = +0.77 V
02

Combine the half-cell reactions to form the overall reaction

Since Fe^3+ gains one electron and Sn^4+ loses two electrons, we need to multiply the second half-cell reaction by 2 to balance the number of electrons: 1. Sn^4+ + 2 e⁻ → Sn^2+; E° = 0.15 V 2. 2(Fe^3+ + e⁻ → Fe^2+); E° = 2 x (+0.77 V) Now, add the two half-cell reactions to form the overall redox reaction: Sn^2+ + 2 Fe^3+ ⇌ Sn^4+ + 2 Fe^2+
03

Calculate the standard electrode potential (E°) of the overall reaction

Now that the overall reaction is balanced, we can calculate the standard electrode potential (E°) using the Nernst equation: E° = E°(reduced) - E°(oxidized) In this equation, E°(reduced) is the standard electrode potential of the reduced species (Fe^3+) and E°(oxidized) is the standard electrode potential of the oxidized species (Sn^4+). E° = (+0.77 V x 2) - 0.15 V = +1.54 V - 0.15 V = +1.39 V
04

Determine if the reaction is spontaneous

Since the standard electrode potential (E°) of the overall reaction is positive (+1.39 V), the reaction Sn^2+ + 2 Fe^3+ ⇌ Sn^4+ + 2 Fe^2+ will proceed spontaneously at 25°C.

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Key Concepts

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

Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical changes. It studies how chemical energy is converted to electrical energy and vice versa in electrochemical cells. These cells are the basis for battery technology, electroplating, and various sensing devices. At the heart of electrochemistry are redox (reduction-oxidation) reactions, where electrons are transferred between substances, leading to a change in oxidation states.

In analyzing redox reactions, electrochemists use a measure known as standard electrode potential, which predicts the direction electrons flow during these reactions. This standard electrode potential is a fundamental concept that helps to determine the feasibility of a reaction and is used in calculations involving the Nernst equation which furthers our understanding of the reaction's behavior under non-standard conditions.
Redox Reactions
Redox reactions are chemical processes where electrons are transferred between two substances, leading to changes in their oxidation states. The substance that loses electrons is oxidized, while the substance that gains electrons is reduced. These reactions are fundamental to various processes in electrochemistry.

To predict the direction and spontaneity of redox reactions, we often refer to standard electrode potentials as a reference. A higher positive value indicates a greater tendency to gain electrons (be reduced), while a lower value suggests a stronger inclination to lose electrons (be oxidized). By comparing the potentials of two half-reactions, we can determine if the overall redox reaction is spontaneous or non-spontaneous.
Nernst Equation
The Nernst equation is a mathematical expression that relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature, and the activities (often approximated by concentrations) of the chemical species involved. It is given by the formula:
\[ E = E^\circ - \frac{RT}{nF} \ln \frac{\text{Products}}{\text{Reactants}} \]
where \(E\) is the cell potential, \(E^\circ\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature in kelvins, \(n\) is the number of moles of electrons transferred, and \(F\) is the Faraday constant. The Nernst equation allows us to calculate the actual electrode potential under non-standard conditions, hence it's vital not only for theoretical calculations but also for practical applications in electrochemical cells and batteries.
Spontaneous Reactions
In the context of redox reactions, a spontaneous reaction is one that occurs naturally without the input of additional energy from the outside. Spontaneity is often gauged by the sign of the standard electrode potential (\(E^\circ\)) for the overall reaction. A positive \(E^\circ\) indicates a spontaneous reaction under standard conditions (298 K, 1 atm, and 1 M concentrations).

This ties into the Gibbs free energy change (\(\Delta G\)), where a negative \(\Delta G\) corresponds to a spontaneous process. The relationship between \(\Delta G\) and \(E^\circ\) can be expressed by the following equation:
\[ \Delta G = -nFE^\circ \]
This equation is a bridge between thermodynamics and electrochemistry, linking the electrical work that can be gained from a redox reaction with its spontaneity. Utilizing these concepts, we can predict not only if the reaction will proceed without the need for external energy but also understand the electrochemical basis of the reaction's favorability.

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Most popular questions from this chapter

You pass a \(1.0\) amp current through an electrolytic cell for \(1.0 \mathrm{hr}\). There are 96,500 coul in a Faraday (F). Calculate the number of grams of each of the following that would be deposited at the cathode: (2) Cu from a \(\mathrm{Cu}^{+2}\) solution and (1) Ag from an Ag ' solution, (3) A.1 from an \(\mathrm{Al}^{3+}\) solution.

Using the tables, of standard electrode potentials, list the fol lowing ions in order of decreasing ability as oxidizing agents: \(\mathrm{Fe}^{3+}, \mathrm{F}_{2}, \mathrm{~Pb}^{2+}, \mathrm{I}_{2}, \mathrm{Sn}^{4+}, \mathrm{O}_{2}\)

Calculate the voltage (E) of a cell with \(\mathrm{E}^{\circ}=1.1\) volts, If the copper half-cell is at standard conditions but the zinc ion concentration is only \(.001\) molar. Temperature is \(25^{\circ} \mathrm{c}\). The overall reaction is \(\mathrm{Zn}+\mathrm{Cu}^{+2} \rightarrow \mathrm{Cu}+\mathrm{Zn}^{+2}\)

You have the following cell process: \(\mathrm{Fe}(\mathrm{s})+\mathrm{Co}^{2+}(.5 \mathrm{M}) \rightarrow \mathrm{Fe}^{2+}(1.0 \mathrm{M})+\mathrm{Co}(\mathrm{s})\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \leftrightarrows \mathrm{Fe}(\mathrm{s})\) with \(\mathrm{E}^{\circ}=-.44 \mathrm{e}\) and \(\mathrm{Co}^{2+}+2 \mathrm{e}^{-} \leftrightharpoons \mathrm{Co}(\mathrm{s})\) with \(\mathrm{E}^{\circ}=-.28\), find the standard cell potential \(\Delta \mathrm{E}\), the cell potential \(\Delta \mathrm{E}\) and the concentration ratio at which the potential generated by the cell is exactly zero. which the potential generated by the cell is exactly zero.

For the following oxidation-reduction reaction, (a) write out the two half- reactions and balance the equation, (b) calculate \(\Delta E^{\circ}\), and (c) determine whether the reaction will proceed spontaneously as written; \(\mathrm{Fe}^{2+}+\mathrm{MnO}^{-}{ }_{4}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (1) \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \leftrightarrows \mathrm{Fe}^{2+}, \mathrm{E}^{\circ}=0.77 \mathrm{eV}\) (2) \(\mathrm{MnO}^{-}{ }_{4}+8 \mathrm{H}^{+}+6 \mathrm{e}^{-} \leftrightarrows \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.51 \mathrm{eV}\)

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