Problem 66 - Chemistry: The Central Science in SI Units, Global Edition

Question

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 \(\left[\mathrm{Cr}^{3+}\right]=1.5 \mathrm{M} ?\)

Step-by-Step Solution

Answer

(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 1: 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.

Step 2: 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.

Step 3: 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.

Key Concepts

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.