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A cell has a standard cell potential of \(+0.257 \mathrm{~V}\) at \(298 \mathrm{~K}\). What is the value of the equilibrium constant for the reaction \((\mathbf{a})\) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?\) (c) if \(n=3 ?\)

Short Answer

Expert verified
For n=1, \(K \approx 2.74 \times 10^4\); for n=2, \(K \approx 7.51 \times 10^7\); for n=3, \(K \approx 2.06 \times 10^{11}\).

Step by step solution

01

Understanding the Nernst Equation

The Nernst equation relates the standard cell potential \(E^°\) to the equilibrium constant \(K\). The equation is given by \[ E^° = \frac{0.0592}{n} \log K \] at standard temperature (\(298\, \text{K}\)).
02

Rearrange the Equation for K

To find \(K\), rearrange the Nernst equation to:\[ \log K = \frac{nE^°}{0.0592} \] Then solve for \(K\):\[ K = 10^{\left(\frac{nE^°}{0.0592}\right)} \].
03

Substitute Values and Solve for n=1

For \(n=1\), substitute \(E^° = 0.257 \mathrm{~V}\) into the rearranged equation:\[ \log K = \frac{1 \times 0.257}{0.0592} \] Calculate this to find \(K\). \[ K = 10^{\left(\frac{0.257}{0.0592}\right)} \approx 2.74 \times 10^4 \].
04

Substitute Values and Solve for n=2

For \(n=2\), the equation becomes:\[ \log K = \frac{2 \times 0.257}{0.0592} \] Calculate to determine \(K\).\[ K = 10^{\left(\frac{0.514}{0.0592}\right)} \approx 7.51 \times 10^7 \].
05

Substitute Values and Solve for n=3

For \(n=3\), the equation becomes:\[ \log K = \frac{3 \times 0.257}{0.0592} \] Calculate to find \(K\).\[ K = 10^{\left(\frac{0.771}{0.0592}\right)} \approx 2.06 \times 10^{11} \].

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Key Concepts

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

Standard Cell Potential
The standard cell potential, represented as \(E^°\), is a crucial concept in electrochemistry. It measures the driving force behind a chemical reaction occurring in an electrochemical cell under standard conditions. Standard conditions generally mean a temperature of 298 K, a pressure of 1 atm, and concentrations of 1 M for any solutions involved.

The standard cell potential is determined by the difference in potentials between the two electrodes in a galvanic cell. It's expressed in volts (V), and it's calculated by subtracting the standard reduction potential of the anode reaction from that of the cathode reaction. An essential detail is that a positive \(E^°\) suggests the reactants spontaneously convert to products, which is a favorable and forward reaction.

To sum it up:
  • Standard conditions: 298 K, 1 atm, 1 M concentrations
  • Positive \(E^°\) implies a spontaneous reaction
  • Measured in volts
Nernst Equation
The Nernst Equation is a fundamental formula in electrochemistry that relates the standard cell potential to the concentration of the involved species. It provides a way to calculate the actual cell potential at any given concentrations, not just under standard conditions. The Nernst Equation is represented as: \[E = E^° - \frac{RT}{nF} \ln{Q}\]where \(E\) is the cell potential at any conditions, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons exchanged, and \(F\) is Faraday's constant.

However, at standard temperature (298 K), this equation simplifies to: \[E = E^° - \frac{0.0592}{n} \log Q\]When a reaction reaches equilibrium, \(E\) becomes zero, and \(Q\) equals the equilibrium constant \(K\). By rearranging the equation, you can solve for \(K\) to understand how far the reaction will proceed.

To highlight its use:
  • Predicts cell potential under non-standard conditions
  • Links standard potential to equilibrium constant \(K\)
  • Helps understand reaction progress
Electrochemistry
Electrochemistry is the branch of chemistry that explores the relationship between electricity and chemical reactions. It involves studying reactions where electrons are transferred, called redox reactions, and these usually occur in electrochemical cells.

Electrochemical cells are devices that convert chemical energy into electrical energy (as in galvanic cells) or use electrical energy to provoke chemical changes (as in electrolytic cells). These principles power various everyday applications—from batteries that power devices to processes like electroplating and electrolysis.

In a practical sense:
  • Galvanic cells transform chemical energy into electrical energy
  • Electrolytic cells drive chemical changes with electrical energy
  • Key reactions involve electron transfer, known as redox reactions
Understanding electrochemistry allows us to harness electrical energy in practical applications, which is essential in science and technology.

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Most popular questions from this chapter

A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an iron strip placed in a solution of \(\mathrm{FeSO}_{4},\) and the other has an aluminum strip placed in a solution of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .\) The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Fe}^{2+}(a q) \longrightarrow 3 \mathrm{Fe}(s)+2 \mathrm{Al}^{3+}(a q) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the iron electrode or from the iron to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the \(\mathrm{Al}\) is not coated with its oxide.

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? [Section 20.5\(]\)

Iron corrodes to produce rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) but other corrosion products that can form are Fe(O)(OH), iron oxyhydroxide, and magnetite, \(\mathrm{Fe}_{3} \mathrm{O}_{4} \cdot(\mathbf{a})\) What is the oxidation number of Fe in iron oxyhydroxide, assuming oxygen's oxidation number is \(-2 ?\) (b) The oxidation number for Fe in magnetite was controversial for a long time. If we assume that oxygen's oxidation number is -2 , and Fe has a unique oxidation number, what is the oxidation number for Fe in magnetite? (c) It turns out that there are two different kinds of Fe in magnetite that have different oxidation numbers. Suggest what these oxidation numbers are and what their relative stoichiometry must be, assuming oxygen's oxidation number is -2 .

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. (a) \(14 \mathrm{H}^{+}(a q)+2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{NaBiO}_{3}(s)\) $$ \longrightarrow 7 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{Bi}^{3+}(a q)+5 \mathrm{Na}^{+}(a q) $$ (b) \(2 \mathrm{KMnO}_{4}(a q)+3 \mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) $$ \longrightarrow 2 \mathrm{MnO}_{2}(s)+3 \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{KOH}(a q) $$ (c) \(\mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. $$ \begin{array}{l} \text { (a) } 2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{HNO}_{3}(a q) \\ \text { (b) } \mathrm{FeS}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g) \\ \text { (c) } \mathrm{Fe}(s)+2 \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+ \end{array} $$ \(2 \mathrm{NO}_{2}(g)+\mathrm{FeO}(s)\)

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