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Chapter 20: Problem 57

(a) Under what circumstances is the Nernst equation applicable? (b) What is the numerical value of the reaction quotient, \(Q\), under standard conditions? (c) What happens to the emf of a cell if the concentrations of the reactants are increased?

Short Answer

Expert verified
(a) The Nernst equation is applicable when the involved species are in equilibrium, the temperature is reasonably constant, and the reaction involves electron transfer (redox reaction). It is given by: \(E = E^\circ - \dfrac{RT}{nF} \ln Q_\mathrm{r}\). (b) Under standard conditions, the numerical value of the reaction quotient, \(Q\), is 1. (c) When the concentrations of the reactants are increased, the cell potential (emf) will also increase, indicating that the cell can provide more electrical energy with higher concentrations of reactants.

Step by step solution

01

(a) The Nernst equation and its applicability

The Nernst equation is used to determine the reduction potential of a half-cell in an electrochemical cell, taking into account the concentration or partial pressure of the involved species. It can be written as: \(E = E^\circ - \dfrac{RT}{nF} \ln Q_\mathrm{r}\) Where: - \(E\) is the cell potential (or emf) at non-standard conditions - \(E^\circ\) is the cell potential at standard conditions - \(R\) is the gas constant (\(8.314\, \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}\)) - \(T\) is the temperature in Kelvin - \(n\) is the number of moles of electrons transferred in the redox reaction - \(F\) is the Faraday's constant (\(9.6485 \times 10^4\, \mathrm{C} \mathrm{mol}^{-1}\)) - \(Q_\mathrm{r}\) is the reaction quotient The Nernst equation is applicable when: 1. The involved species are in equilibrium. 2. The temperature is reasonably constant. 3. The reaction involves electron transfer (redox reaction).
02

(b) Reaction quotient under standard conditions

Under standard conditions, the concentration of each species is 1 M, and the partial pressure of the involved gases is 1 atm. The reaction quotient Q can be defined as: \(Q = \dfrac{\prod([\text{products}]^{\text{stoichiometric coefficients}})}{\prod([\text{reactants}]^{\text{stoichiometric coefficients}})}\) Under standard conditions, the concentrations of all species are the same (1 M). Therefore, the reaction quotient Q becomes: \(Q = \dfrac{1^{\text{stoichiometric coefficients of products}}}{1^{\text{stoichiometric coefficients of reactants}}}\) Since any number raised to any power is always one, the numerical value of Q under standard conditions is: \(Q = 1\)
03

(c) Effect of increasing reactants concentrations on emf

According to the Nernst equation, the cell potential is related to the natural logarithm of the reaction quotient Q: \(E = E^\circ - \dfrac{RT}{nF} \ln Q_\mathrm{r}\) When the concentrations of the reactants are increased, the reaction quotient Q decreases (less than 1) because Q is the ratio of the concentrations of products and reactants. As a result, the value of the natural logarithm \(\ln Q_\mathrm{r}\) becomes negative. This causes the second term (\(\dfrac{RT}{nF} \ln Q_\mathrm{r}\)) in the Nernst equation to become positive. Therefore, as the concentrations of the reactants increase, the cell potential (emf) will also increase. This indicates that the cell can provide more electrical energy when the concentrations of reactants are increased.

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Most popular questions from this chapter

(a) The nonrechargeable lithium batteries used for photography use lithium metal as the anode. What advantages might be realized by using lithium rather than zinc, cadmium, lead, or nickel? (b) The rechargeable lithiumion battery does not use lithium metal as an electrode material. Nevertheless, it still has a substantial advantage over nickel-based batteries. Suggest an explanation.

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+& 4 \mathrm{Br}^{-}(a q) \\ & E_{\text {red }}^{\circ}=-0.858 \mathrm{~V} \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & \\\ E_{\text {red }}^{\circ}=-0.43 \mathrm{~V} \end{aligned} $$ $$ \begin{aligned} \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) &+2 \mathrm{OH}^{-}(a q) \\ & E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \\ \mathrm{Sn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{Sn}(s) & \\\ E_{\mathrm{red}}^{\circ}=-0.14 \mathrm{~V} \end{aligned} $$ (a) Write the cell reaction for the combination of these half-cell reactions that leads to the largest positive cell emf, and calculate the value. (b) Write the cell reaction for the combination of half-cell reactions that leads to the smallest positive cell emf, and calculate that value.

A voltaic cell is constructed that uses the following halfcell reactions: $$ \begin{gathered} \mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{gathered} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=3.5 \mathrm{M}\). (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If \(\left[\mathrm{Cu}^{+}\right]\) was equal to \(0.15 \mathrm{M}\), at what concentration of \(\mathrm{I}^{-}\) would the cell have zero potential?

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

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{3}\), and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\). The overall cell reaction is $$ \mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two electrode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode, or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?
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