Predict whether the following reactions would occur spontaneously in aqueous solution at \(25^{\circ} \mathrm{C}\). Assume that the initial concentrations of dissolved species are all \(1.0 M\) (a) \(\mathrm{Ca}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{Cd}(s)\) (b) \(2 \mathrm{Br}^{-}(a q)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Br}_{2}(l)+\operatorname{Sn}(s)\) (c) \(2 \mathrm{Ag}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Ni}(s)\) (d) \(\mathrm{Cu}^{+}(a q)+\mathrm{Fe}^{3+}(a q) \longrightarrow\) $$\mathrm{Cu}^{2+}(a q)+\mathrm{Fe}^{2+}(a q)$$

Short Answer

Expert verified
Reactions (a) and (d) would occur spontaneously while reactions (b) and (c) would not.

Step by step solution

01

Understanding the Task

We need to predict the spontaneity of the given reactions. According to the Nernst equation, for a reaction to occur spontaneously, the cell potential (\(E_{cell}\)) should be greater than zero.
02

Checking the Standard Reduction Potential Table

Look at the standard reduction potential table. This table provides a list of the potential differences of all elements. The elements are arranged in the order of their potential to gain electrons (it's a measure of how good an oxidizing agent they are). For the given reaction, if the reducing species has a higher reduction potential than the oxidizing species, the reaction will be spontaneous. This is because a higher reduction potential means more tendency to gain electrons (be reduced), which would push the reaction forward.
03

Case (a)

For the reaction \(\mathrm{Ca}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{Cd}(s)\), Calcium is getting oxidized and Cadmium is getting reduced. From Standard Reduction Potential table, reduction potential of \(Ca^{2+}/Ca\) is -2.87 V and \(Cd^{2+}/Cd\) is -0.40 V. Since reduction potential of reactant (which gets oxidized) is less than the product (which gets reduced), the reaction is spontaneous.
04

Case (b)

For the reaction \( 2 \mathrm{Br}^{-}(a q)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Br}_{2}(l)+\operatorname{Sn}(s)\), \(Br^{-}\) is getting oxidized and \(Sn^{2+}\) is getting reduced. From Standard Reduction Potential table, reduction potential of \(Br_{2}/Br^{-}\) is 1.087 V and for \(Sn^{2+}/Sn\) is -0.136 V. Since reduction potential of reactant (which gets oxidized) is more than the product (which gets reduced), the reaction is not spontaneous.
05

Case (c)

For the reaction \(2 \mathrm{Ag}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Ni}(s)\), \(Ag\) is getting oxidized and \(Ni^{2+}\) is getting reduced. From Standard Reduction Potential table, reduction potential of \(Ag/A^{+}\) is 0.7996 V and for \(Ni^{2+}/Ni\) is -0.257 V. Since reduction potential of reactant (which gets oxidized) is more than the product (which gets reduced), the reaction is not spontaneous.
06

Case (d)

For the reaction \(\mathrm{Cu}^{+}(a q)+\mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Fe}^{2+}(a q)\), \(Cu^{+}\) is getting oxidized and \(Fe^{3+}\) is getting reduced. From Standard Reduction Potential table, reduction potential of \(Cu^{2+}/Cu^{+}\) is 0.153 V and for \(Fe^{3+}/Fe^{2+}\) is 0.771 V. Since reduction potential of reactant (which gets oxidized) is less than the product (which gets reduced), the reaction is spontaneous.

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Key Concepts

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

Standard Reduction Potential
The Standard Reduction Potential (SRP) is an intrinsic property of a substance that measures its tendency to gain electrons and be reduced. It is defined with respect to a standard hydrogen electrode, which has an assigned potential of 0 volts. Chemical species with more positive SRP are better oxidizing agents, meaning they prefer to gain electrons, while species with more negative SRP are better reducing agents, willing to donate electrons.

When analyzing redox reactions, comparing SRP values helps determine the direction in which electrons will flow. The more positive the SRP of a reactant, the greater its ability to be reduced. If a reacting species has a higher SRP than another, it is more likely to be reduced, pulling electrons away from the other substance, which gets oxidized.
Nernst Equation
The Nernst equation is foundational in understanding electrochemical spontaneity. It quantifies the cell potential at any concentration, giving insight beyond the standard conditions considered by SRP. The Nernst equation is expressed as:

\( E = E^0 - \frac{RT}{nF} \ln(Q) \),

where \(E\) is the cell potential, \(E^0\) the standard cell potential, \(R\) the ideal gas constant, \(T\) the temperature in Kelvin, \(n\) the number of electrons transferred, \(F\) the Faraday's constant, and \(Q\) the reaction quotient. For spontaneous reactions at standard conditions, the cell potential calculated using the Nernst equation must be greater than zero. It allows predictions about reactions' behavior under varying concentrations, enhancing the understanding beyond what SRP alone can provide.
Electrochemical Spontaneity
Electrochemical spontaneity refers to the natural tendency of a redox reaction to occur without the input of external energy. It is determined by the cell potential, which arises from the difference in SRP values of the reactant and product couples. A reaction is spontaneous if the overall cell potential is positive, as indicated by both the standard reduction potentials and the Nernst equation.

A positive cell potential indicates that the driving force for the transfer of electrons is favorable. In such cases, the reaction can release energy, usually as electrical work in a galvanic cell. Conversely, if the cell potential is negative, the reaction is non-spontaneous under standard conditions and would require external energy to proceed, such as in an electrolytic cell.
Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions are chemical processes involving the transfer of electrons from one species to another. The species that loses electrons is oxidized, while the one that gains electrons is reduced. Working out whether these reactions are spontaneous requires an understanding of the involved species' redox potentials.

In a redox reaction context, such as the given exercises, it's crucial to identify the oxidizing and reducing agents. The reactant being oxidized must have a less positive, or more negative, SRP than the reactant being reduced. In other words, the oxidizing agent should have a greater tendency to gain electrons (higher SRP) compared with the reducing agent. When the oxidizing agent has a lower SRP value than the reducing agent, the reaction is not spontaneous and would not proceed under standard conditions.

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Most popular questions from this chapter

Balance the following redox equations by the ionelectron method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution)

Which species in each pair is a better oxidizing agent under standard-state conditions: (a) \(\mathrm{Br}_{2}\) or \(\mathrm{Au}^{3+} ?\) (b) \(\mathrm{H}_{2}\) or \(\mathrm{Ag}^{+} ?\) (c) \(\mathrm{Cd}^{2+}\) or \(\mathrm{Cr}^{3+} ?\) (d) \(\mathrm{O}_{2}\) in acidic media or \(\mathrm{O}_{2}\) in basic media?

The passage of a current of 0.750 A for \(25.0 \mathrm{~min}\) deposited \(0.369 \mathrm{~g}\) of copper from a \(\mathrm{CuSO}_{4}\) solution. From this information, calculate the molar mass of copper.

Describe the electrolysis of an aqueous solution of \(\mathrm{KNO}_{3}\)

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) at \(25^{\circ} \mathrm{C}\). The measured potential difference between the rod and the \(\mathrm{SHE}\) is \(0.589 \mathrm{~V},\) the rod being positive. Calculate the solubility product constant for silver oxalate.

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