Question

A voltaic cell utilizes the following reaction and operates at 298 K: $$3{\mathrm{Ce}}^{4+}(aq)+\mathrm{Cr}(s)⟶3{\mathrm{Ce}}^{3+}(aq)+{\mathrm{Cr}}^{3+}(aq)$$ (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

(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

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: $$\begin{array}{rl}& \text{Anode: }3{\mathrm{Ce}}^{3+}(\mathrm{aq})\to 3{\mathrm{Ce}}^{4+}(\mathrm{aq})+3e^{-}(E_{{\mathrm{Ce}}^{3+}/{\mathrm{Ce}}^{4+}}^{\circ })\\ & \text{Cathode: }{\mathrm{Cr}}^{3+}(\mathrm{aq})+3e^{-}\to \mathrm{Cr}(\mathrm{s})(E_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\circ })\end{array}$$ Using a standard reduction potential table, we find: $E_{{\mathrm{Ce}}^{3+}/{\mathrm{Ce}}^{4+}}^{\circ }=1.61\mathrm{V}$ and $E_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\circ }=-0.74\mathrm{V}$ 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.

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: $$[{\mathrm{Ce}}^{4+}]=3.0\mathrm{M},[{\mathrm{Ce}}^{3+}]=0.10\mathrm{M},[{\mathrm{Cr}}^{3+}]=0.010\mathrm{M}$$ 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.

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: ${\mathrm{Cr}}^{3+}(\mathrm{aq})+3e^{-}\to \mathrm{Cr}(\mathrm{s})$- Anode: $3{\mathrm{Ce}}^{3+}(\mathrm{aq})\to 3{\mathrm{Ce}}^{4+}(\mathrm{aq})+3e^{-}$The standard reduction potentials are:

• $E_{{\mathrm{Ce}}^{3+}/{\mathrm{Ce}}^{4+}}^{\circ }=1.61\mathrm{V}$
• $E_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\circ }=-0.74\mathrm{V}$

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:- ${\mathrm{Ce}}^{4+}/{\mathrm{Ce}}^{3+}$ with $E^{\circ }=1.61\mathrm{V}$ suggests it's a strong oxidizing agent, favoring reduction.- ${\mathrm{Cr}}^{3+}/\mathrm{Cr}$ 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 $(8.314\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})$.
• $T$ is the temperature in Kelvin.
• $n$ is the number of moles of electrons exchanged.
• $F$ is Faraday's constant $(96485\mathrm{C}/\mathrm{mol})$.
• $Q$ 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.