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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

Expert verified
(a) Under standard conditions, the emf of the voltaic cell is \(0.46 \mathrm{V}\). (b) Under the given non-standard conditions, the emf of the voltaic cell is \(0.832 \mathrm{V}\).

Step by step solution

01

Write the balanced equation for the voltaic cell reaction

The given reaction is already balanced: \[ 4 \mathrm{Fe}^{2+}(aq)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(aq) \rightarrow 4 \mathrm{Fe}^{3+}(aq)+2 \mathrm{H}_{2} \mathrm{O}(l) \]
02

Calculate the emf under standard conditions

To calculate the emf under standard conditions, we'll use the standard reduction potential values. The overall reaction can be divided into two half-reactions: 1. Oxidation half-reaction: \(\mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}^{3+}(aq) + e^-\) 2. Reduction half-reaction: \(\mathrm{O}_2(g) + 4 \mathrm{H}^+ + 4e^- \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)\) The standard reduction potential values for these half-reactions are: 1. \(E^0_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 0.77 \mathrm{V}\) 2. \(E^0_{\mathrm{O}_2/\mathrm{H}_2 \mathrm{O}} = 1.23 \mathrm{V}\) Now we can calculate the emf under standard conditions using the formula: \[ \Delta E^0_{cell} = E^0_{cathode} - E^0_{anode} \] In this case, the cathode is the reduction half-reaction, and the anode is the oxidation half-reaction: \[ \Delta E^0_{cell} = 1.23 \mathrm{V} - 0.77 \mathrm{V} = 0.46 \mathrm{V} \] So, under standard conditions, the emf of this cell is 0.46 V.
03

Calculate the emf under non-standard conditions

To calculate the emf under non-standard conditions, we'll use the Nernst equation: \[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \cdot \ln Q \] Here, R is the gas constant (8.314 J/mol·K), T is the temperature (298 K, assuming standard conditions), n is the number of moles of electrons transferred (which is 4 in this case, from the balanced equation), and F is Faraday's constant (96,485 C/mol). Now, let's calculate the reaction quotient Q using the given concentrations and pressure: \[ Q = \frac{\left[\mathrm{Fe}^{3+}\right]^4}{\left[\mathrm{Fe}^{2+}\right]^4 \cdot P_{\mathrm{O}_2} \cdot \left[\mathrm{H}^+\right]^4} = \frac{(0.010)^4}{(1.3)^4 \cdot 0.50 \cdot (0.00316)^4} \] Plug the values into the Nernst equation and solve for E_cell: \[ E_{cell} = 0.46 \mathrm{V} - \frac{(8.314)(298)}{(4)(96485)} \cdot \ln Q = 0.832 \mathrm{V} \] So, under the given non-standard conditions, the emf of this cell is 0.832 V.

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Key Concepts

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

Standard Reduction Potential
Understanding the standard reduction potential is essential in electrochemistry, as it measures the tendency of a chemical species to acquire electrons and be reduced. This value, expressed in volts (V), is determined under standard conditions, which are a concentration of 1 M for aqueous solutions, a pressure of 1 atm for gases, and a temperature of 25°C (298 K).

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
The Nernst equation plays a pivotal role in determining the EMF of an electrochemical cell under non-standard conditions. It takes into consideration the effect of ion concentration, partial pressure of gases, and temperature on cell potential. The Nernst equation is mathematically represented as:

\[\begin{equation}E_{cell} = E^0_{cell} - \frac{RT}{nF} \cdot \ln Q\end{equation}\]
Here, \begin{itemize}
  • E_{cell} is the cell potential under non-standard conditions,
  • E^0_{cell} is the standard EMF of the cell,
  • R is the ideal gas constant,
  • T is the temperature in kelvins,
  • n is the number of moles of electrons transferred in the redox reaction,
  • F is Faraday's constant, and
  • Q is the reaction quotient, which expresses the ratio of products to reactants at a given moment.

  • 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
    The reaction quotient (Q) is a measure used in chemistry to determine the direction in which a reaction is proceeding at any given moment. Q is calculated in the same way as the equilibrium constant (K), using the concentrations of the reactants and the products raised to the power of their coefficients in the balanced equation; however, unlike K, the concentrations are not necessarily those at equilibrium.

    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
    The field of electrochemistry is at the intersection of electricity and chemistry, involving processes where chemical reactions create electrical currents or vice versa. Voltaic cells, also known as galvanic cells, are a fundamental concept within this field, as they convert chemical energy into electrical energy through spontaneous redox reactions.

    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.

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    Most popular questions from this chapter

    (a) A voltaic cell is constructed with all reactants and products in their standard states. Will this condition hold as the cell operates? Explain. (b) Can the Nernst equation be used at temperatures other than room temperature? Explain. (c) What happens to the emf of a cell if the concentrations of the products are increased?

    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?

    The hydrogen-oxygen fuel cell has a standard emf of \(1.23 \mathrm{~V}\). What advantages and disadvantages are there to using this device as a source of power, compared to a 1.55-V alkaline battery?

    (a) Why is it impossible to measure the standard reduction potential of a single half-reaction? (b) Describe how the standard reduction potential of a half-reaction can be determined.

    (a) Based on standard reduction potentials, would you expect copper metal to oxidize under standard conditions in the presence of oxygen and hydrogen ions? (b) When the Statue of Liberty was refurbished, Teflon spacers were placed between the iron skeleton and the copper metal on the surface of the statue. What role do these spacers play?

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