Chapter 20: Problem 63
A voltaic cell utilizes the following reaction: $$ 4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)-\cdots \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M}\), \(\left[\mathrm{Fe}^{3+}\right]=0.010 \mathrm{M}, P_{\mathrm{O}_{2}}=0.50 \mathrm{~atm}\), and the \(\mathrm{pH}\) of the so- lution in the cathode compartment is \(3.50 ?\)
Short Answer
Step by step solution
Write the balanced equation for the voltaic cell reaction
Calculate the emf under standard conditions
Calculate the emf under non-standard conditions
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Reduction Potential
The more positive the standard reduction potential, the greater the species' affinity for electrons and the stronger its oxidizing power. Conversely, a more negative value indicates a weaker tendency to gain electrons and a stronger reducing power. In a voltaic cell, two half-reactions occur - one at the cathode (reduction) and one at the anode (oxidation). By comparing the standard reduction potentials of the two half-reactions, we can determine the direction of electron flow and calculate the standard electromotive force (EMF) of the cell, which is the potential difference between the electrodes when no current is flowing.
In the exercise provided, the standard reduction potentials for the iron and oxygen half-reactions are given as 0.77 V and 1.23 V, respectively. These values allow us to predict that electrons will flow from the iron to the oxygen, resulting in the generation of an electrical current and the ability to do work.
Nernst Equation
\[\begin{equation}E_{cell} = E^0_{cell} - \frac{RT}{nF} \cdot \ln Q\end{equation}\]
Here, \begin{itemize}
For the voltaic cell in the exercise, by applying the Nernst equation with the given values, we can determine the EMF when the cell operates under the specified non-standard conditions.
Reaction Quotient
For a general reaction \[\begin{equation}a A + b B \rightarrow c C + d D\end{equation}\]
the reaction quotient is expressed as:\[\begin{equation}Q = \frac{[C]^c \cdot [D]^d}{[A]^a \cdot [B]^b}\end{equation}\]
Where [A], [B], [C], and [D] represent the molar concentrations of the reactants and products. When Q = K, the system is at equilibrium, meaning no net change occurs. If Q < K, the reaction will proceed forward, and if Q > K, the reaction will go in reverse to achieve equilibrium.
In the given exercise, the reaction quotient helps in applying the Nernst equation to find the EMF under non-standard conditions, thereby accounting for the current concentration of ions and partial pressures involved in the cell's reaction.
Electrochemistry
In a voltaic cell, two different metals are immersed in solutions containing their ions and connected by a wire and a salt bridge. One metal serves as the anode where oxidation occurs and releases electrons, while the other serves as the cathode where reduction occurs and consumes electrons. This flow of electrons through the wire generates an electric current that can be harnessed to do work.
The EMF produced by a voltaic cell can be predicted and calculated using principles from electrochemistry, including the standard reduction potential of the reactions involved and the Nernst equation for non-standard conditions. The understanding of such cells paves the way for developing batteries and other technologies crucial for energy storage and conversion. The voltaic cell exemplified in the exercise is a practical demonstration of using electrochemical principles to compute EMF for both standard and non-standard conditions.