/*! 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 66 A voltaic cell utilizes the foll... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 66

A voltaic cell utilizes the following reaction and operates at 298 K: $$ 3 \mathrm{Ce}^{4+}(a q)+\mathrm{Cr}(s) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ce}^{4+}\right]=3.0 \mathrm{M},\) \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M},\) and \(\left[\mathrm{Cr}^{3+}\right]=0.010 \mathrm{M} ?(\mathbf{c})\) What is the emf of the cell when \(\left[\mathrm{Ce}^{4^{+}}\right]=0.010 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=2.0 \mathrm{M}\) and \(\left[\mathrm{Cr}^{3+}\right]=1.5 \mathrm{M} ?\)

Short Answer

Expert verified
(a) The emf of the cell under standard conditions is -2.35 V. (b) The emf of the cell with the given concentrations (\([\mathrm{Ce^{4+}}]= 3.0\; \mathrm{M},\; [\mathrm{Ce^{3+}}] = 0.10\; \mathrm{M},\; [\mathrm{Cr^{3+}}] = 0.010\; \mathrm{M}\)) is approximately -1.99 V. (c) The emf of the cell with the given concentrations (\(\left[\mathrm{Ce}^{4^{+}}\right]=0.010 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=2.0 \mathrm{M}\) and \(\left[\mathrm{Cr}^{3+}\right]=1.5 \mathrm{M}\)) is approximately -2.51 V.

Step by step solution

01

Calculate emf under standard conditions

To calculate the emf of the cell under standard conditions, we will use the equation: \(E_{cell} = E_{cathode} - E_{anode}\) First, we need to look up the standard electrode potentials for the half-reactions. From the given equation, the half-cell reactions are: \[ \begin{aligned} &\text{Anode: } 3\,\mathrm{Ce^{3+}(aq)} \rightarrow 3\,\mathrm{Ce^{4+}(aq)} + 3\, e^- \quad (E^\circ_{\mathrm{Ce^{3+}/Ce^{4+}}}) \\ &\text{Cathode: } \mathrm{Cr^{3+}(aq)} + 3\, e^- \rightarrow \mathrm{Cr(s)} \quad (E^\circ_{\mathrm{Cr^{3+}/Cr}}) \end{aligned} \] Using a standard reduction potential table, we find: \(E^\circ_{\mathrm{Ce^{3+}/Ce^{4+}}} = 1.61 \,\mathrm{V}\) and \(E^\circ_{\mathrm{Cr^{3+}/Cr}} = -0.74\, \mathrm{V}\) Now, we can calculate the emf under standard conditions: \(E_{cell} = E_{cathode} - E_{anode} = (-0.74\, \mathrm{V}) - (1.61\, \mathrm{V}) = -2.35\, \mathrm{V}\) (a) The emf of the cell under standard conditions is -2.35 V.
02

Calculate emf under given concentrations using Nernst equation

We will now use the Nernst equation to find the emf of the cell under the given concentrations: \(E = E^\circ - \frac{RT}{nF} \ln Q\) where: - \(E\) is the cell potential - \(E^\circ\) is the standard cell potential - \(R\) is the universal gas constant - \(T\) is the temperature (in Kelvin) - \(n\) is the number of moles of electrons transferred in the reaction - \(F\) is Faraday's constant - \(Q\) is the reaction quotient For (b), the concentrations are as follows: \[\mathrm{[Ce^{4+}]}= 3.0\; \mathrm{M},\quad [\mathrm{Ce^{3+}}] = 0.10\; \mathrm{M},\quad [\mathrm{Cr^{3+}}] = 0.010\; \mathrm{M}\] The reaction quotient for this reaction is: \(Q = \frac{[\mathrm{Ce^{4+}]}^3[\mathrm{Cr^{3+}]} }{[\mathrm{Ce^{3+}]}^3} \) Plug in the given values and calculate Q: \(Q = \frac{(3.0)^3 \cdot (0.010)}{(0.10)^3} = 900\) Now calculate the emf for this condition: \(E = -2.35 - \frac{(8.314)(298)}{(3)(96485)} \ln (900)\) \(E = -2.35 - 0.0159 \ln(900) \approx -1.99\; \mathrm{V}\) (b) The emf of the cell with the given concentrations is approximately -1.99 V.
03

Calculate emf under different concentrations using Nernst equation

For (c), the concentrations are as follows: \(\left[\mathrm{Ce}^{4^{+}}\right]=0.010 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=2.0 \mathrm{M} \text{ and }\left[\mathrm{Cr}^{3+}\right]=1.5 \mathrm{M}\) Calculate the reaction quotient for these concentrations: \(Q = \frac{(0.010)^3 \cdot (1.5)}{(2.0)^3} = 0.0009375\) Calculate the emf using the Nernst equation: \(E = -2.35 - \frac{(8.314)(298)}{(3)(96485)} \ln (0.0009375)\) \(E = -2.35 - 0.0159 \ln(0.0009375) \approx -2.51\; \mathrm{V}\) (c) The emf of the cell with the given concentrations is approximately -2.51 V.

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Key Concepts

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

emf calculation
In a voltaic cell, the electromotive force (emf) is a key measure that represents the potential difference between two electrodes. To calculate the emf of a cell under standard conditions, we use the formula: \[ E_{cell} = E_{cathode} - E_{anode} \] where the terms \( E_{cathode} \) and \( E_{anode} \) refer to the standard electrode potentials of the cathode and anode, respectively. The values for these potentials are typically found in standard reduction potential tables.For our given reaction:- Cathode: \( \mathrm{Cr^{3+}(aq)} + 3\, e^- \rightarrow \mathrm{Cr(s)} \)- Anode: \( 3\,\mathrm{Ce^{3+}(aq)} \rightarrow 3\,\mathrm{Ce^{4+}(aq)} + 3\, e^- \)The standard reduction potentials are:
  • \( E^\circ_{\mathrm{Ce^{3+}/Ce^{4+}}} = 1.61 \,\mathrm{V} \)
  • \( E^\circ_{\mathrm{Cr^{3+}/Cr}} = -0.74\, \mathrm{V} \)
Using the formula:\[ E_{cell} = (-0.74) - (1.61) = -2.35\, \mathrm{V} \]This calculation indicates that the emf of the cell under standard conditions is \(-2.35\, \mathrm{V}\). The negative sign suggests a non-spontaneous reaction under these conditions.
standard electrode potentials
Standard electrode potentials are critical in determining how a voltaic cell functions. Each half-reaction in the electrochemical cell is assigned a standard electrode potential \( E^\circ \), measured in volts. This value, derived under standard conditions (1 M concentration, 1 atm pressure, and 298 K), informs us about the tendency of a species to either gain or lose electrons.A higher (more positive) standard electrode potential means a greater likelihood of reduction. Conversely, a lower (more negative) potential indicates a tendency to be oxidized. In our example:- \( \mathrm{Ce^{4+}/Ce^{3+}} \) with \( E^\circ = 1.61 \, \mathrm{V} \) suggests it's a strong oxidizing agent, favoring reduction.- \( \mathrm{Cr^{3+}/Cr} \) with \( E^\circ = -0.74 \, \mathrm{V} \) indicates a stronger tendency for oxidation.These potentials aid in calculating the cell's overall emf, as well as predicting which half-reaction will act as the anode and which will serve as the cathode in a given electrochemical cell.
Nernst equation
The Nernst equation provides a way to calculate the emf of a cell under non-standard conditions, accounting for ion concentrations. It can predict the effect of varying concentrations on cell voltage:\[ E = E^\circ - \frac{RT}{nF} \ln Q \]where:
  • \( E \) is the cell potential under non-standard conditions.
  • \( R \) is the universal gas constant \( (8.314 \, \mathrm{J/mol\cdot K}) \).
  • \( T \) is the temperature in Kelvin.
  • \( n \) is the number of moles of electrons exchanged.
  • \( F \) is Faraday's constant \( (96485 \, \mathrm{C/mol}) \).
  • \( Q \) is the reaction quotient, calculated from the concentrations of the reactants and products.
In our specific problem, when concentrations differ from one mole per liter, the calculated \( Q \) affects \( E \), altering the potential:- For concentrations \( [\mathrm{Ce^{4+}}]=3.0 \mathrm{M}, [\mathrm{Ce^{3+}}]=0.1 \mathrm{M}, \text{and} [\mathrm{Cr^{3+}}]=0.01 \mathrm{M} \), \( Q = 900 \), and the adjusted \( E \) is around \(-1.99\, \mathrm{V}\).- Differing concentrations of \( [\mathrm{Ce^{4+}}]=0.01 \mathrm{M}, [\mathrm{Ce^{3+}}]=2.0 \mathrm{M}, \text{and} [\mathrm{Cr^{3+}}]=1.5 \mathrm{M} \) result in \( Q = 0.0009375 \), shifting \( E \) to approximately \(-2.51\, \mathrm{V}\).This versatile equation is vital for applications where concentrations deviate from the ideal, allowing accurate predictions of electrochemical behavior.

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Most popular questions from this chapter

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q)\) (b) \(\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)\) (acidic solution) (c) \(\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cr}(s)\) (acidic solution) (d) \(\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution)

During the discharge of an alkaline battery, \(4.50 \mathrm{~g}\) of \(\mathrm{Zn}\) is consumed at the anode of the battery. (a) What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from \(\mathrm{Zn}\) to \(\mathrm{MnO}_{2} ?\)

Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution: $$ \begin{array}{lr} \hline \text { Reduction Half-Reaction } & {E^{\circ}(\mathrm{V})} \\ \hline \mathrm{A}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{A}(s) & 1.33 \\\ \mathrm{~B}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{B}(s) & 0.87 \\\ \mathrm{C}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{C}^{2+}(a q) & -0.12 \\ \mathrm{D}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{D}(s) & -1.59 \\\ \hline \end{array} $$ (a) Which substance is the strongest oxidizing agent? Which is weakest? (b) Which substance is the strongest reducing agent? Which is weakest? (c) Which substance(s) can oxidize \(\mathrm{C}^{2+} ?\)

A mixture of copper and gold metals that is subjected to electrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common oxidation state, \(\mathrm{Te}^{4+}\), is $$ \mathrm{Te}^{4+}(a q)+4 \mathrm{e}^{-} \longrightarrow \mathrm{Te}(s) \quad E_{\mathrm{red}}^{\circ}=0.57 \mathrm{~V} $$ Given this information, describe the probable fate of tellurium impurities during electrorefining. Do the impurities fall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If they go into solution, do they plate out on the cathode?

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l} \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{array} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text {red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?
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