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A voltaic cell is constructed that uses the following reaction and operates at \(298 \mathrm{~K}\) : $$\mathrm{Zn}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Ni}(s)$$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ni}^{2+}\right]=3.00 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ni}^{2+}\right]=0.200 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)

Short Answer

Expert verified
The emf of the cell under the given conditions are: (a) 0.53 V under standard conditions (b) 0.574 V when \([\text{Ni}^{2+}]=3.00 \text{ M}\) and \([\text{Zn}^{2+}]=0.100 \text{ M}\) (c) 0.511 V when \([\text{Ni}^{2+}]=0.200 \text{ M}\) and \([\text{Zn}^{2+}]=0.900 \text{ M}\)

Step by step solution

01

Identify Half-cell Reactions and Standard Reduction Potentials

First, let's break down the given cell reaction into half-cell reactions: Zn(s) 鉄 Zn虏鈦(aq) + 2e鈦 (oxidation half-reaction) Ni虏鈦(aq) + 2e鈦 鉄 Ni(s) (reduction half-reaction) Now we need to obtain the standard reduction potentials (E掳) for these half-cell reactions from a standard reduction potential table: E掳(Zn虏鈦/Zn) = -0.76 V E掳(Ni虏鈦/Ni) = -0.23 V
02

Apply Nernst Equation

The Nernst equation helps us to find the emf of the cell under non-standard conditions: \(E = E掳 - \frac{RT}{nF} \times \ln{Q}\) where E = emf of the voltaic cell E掳 = standard emf R = gas constant 8.314 J/(mol路K) T = temperature in Kelvin (298 K) n = number of moles of electrons transferred in the balanced cell reaction (2 moles in this case) F = Faraday's constant 96485 C/mol Q = reaction quotient (equal to [Zn虏鈦篯/[Ni虏鈦篯)
03

Find the Emf under Standard Conditions

For a cell reaction under standard conditions, the emf is the difference in the reduction potentials of the two half-cell reactions: E掳 = E掳(Ni虏鈦/Ni) - E掳(Zn虏鈦/Zn) E掳 = (-0.23 V) - (-0.76 V) E掳 = 0.53 V
04

Calculate the Emf for the given Concentrations

To calculate the emf of the cell when [Ni虏鈦篯=3.00 M and [Zn虏鈦篯=0.100 M, we need to plug these values and the calculated E掳 into the Nernst equation: E = 0.53 V - ((8.314 J/(mol K))*(298 K))/(2*96485 C/mol) * ln(0.100/3.00) E = 0.53 V - 0.0129 * ln(0.0333) E = 0.53 V - 0.0129 * (-3.409) E = 0.53 V + 0.044 E = 0.574 V
05

Calculate the Emf for the given Concentrations

Now we will calculate the emf of the cell when [Ni虏鈦篯=0.200 M and [Zn虏鈦篯=0.900 M: E = 0.53 V - ((8.314 J/(mol K))*(298 K))/(2*96485 C/mol) * ln(0.900/0.200) E = 0.53 V - 0.0129 * ln(4.5) E = 0.53 V - 0.0129 * 1.504 E = 0.53 V - 0.019 E = 0.511 V So, the emf of the cell under the given conditions are: (a) 0.53 V under standard conditions (b) 0.574 V when [Ni虏鈦篯=3.00 M and [Zn虏鈦篯=0.100 M (c) 0.511 V when [Ni虏鈦篯=0.200 M and [Zn虏鈦篯=0.900 M

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Key Concepts

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

Nernst Equation
The Nernst equation is a fundamental tool in electrochemistry, providing a way to calculate the electromotive force (emf) of an electrochemical cell under non-standard conditions. Simply put, this equation allows us to predict how the voltage of a cell will change when the concentrations of the reactants and products vary from their standard state values.

The equation is expressed as:
\[E = E掳 - \frac{RT}{nF} \times \ln{Q}\]
In this formula, \(E\) is the emf of the voltaic cell, \(E掳\) is the standard emf, \(R\) 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, and \(Q\) is the reaction quotient, which reflects the ratio of product concentrations to reactant concentrations.

By applying this equation, we can see how the cell potential is influenced by temperature and the concentration of ions. The presence of the natural logarithm function in the Nernst equation shows that the relationship between concentrations and cell potential is logarithmic, not linear.
Standard Reduction Potentials
Standard reduction potentials, usually denoted as \(E掳\), are a measure of the tendency of a chemical species to gain electrons and thereby be reduced. Each half-cell reaction in an electrochemical cell has its own standard reduction potential, and these can be found in published tables for various reactions.

These standard potentials are measured under standard conditions, which are typically 25掳C (298 K), 1 atm of pressure, and 1 M concentration for each aqueous species involved in the reaction. The more positive the standard reduction potential, the greater the species' affinity for electrons. A species with a highly positive \(E掳\) is likely to be a good oxidizing agent, while one with a negative \(E掳\) would be a good reducing agent.

By comparing the standard reduction potentials of two half-cells, we can predict the direction of electron flow in the cell and calculate the cell's standard emf by subtracting the potential of the anode from that of the cathode. For instance, in the given exercise, the standard emf of the cell is calculated by subtracting the \(E掳\) value of the zinc half-cell from that of the nickel half-cell.
Electrochemical Cell Reactions
Electrochemical cell reactions are the chemical processes that occur at the electrodes of an electrochemical cell, causing the flow of electrons through an external circuit. The cell is composed of two half-cells linked by a conductive material, with each half-cell containing an electrode and an electrolyte.

In a voltaic cell, or galvanic cell, the chemical reaction is spontaneous, driving the flow of electrons from the anode to the cathode. This flow of electrons provides electrical power that can be harnessed for external use. The oxidation half-reaction occurs at the anode, where electrons are released, while reduction occurs at the cathode, where electrons are gained.

Combining the half-reactions provides the overall cell reaction, as seen in the exercise, where zinc solid \( \text{Zn}(s) \) undergoes oxidation and nickel ions \( \text{Ni}^{2+}(aq) \) undergo reduction. The standard emf of the cell, calculated based on standard reduction potentials, indicates the maximum potential difference between electrodes when the concentrations of reactants and products are at standard conditions.

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Most popular questions from this chapter

A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{3}\), and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\). The overall cell reaction is $$\mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s)$$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

(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?

(a) What is electrolysis? (b) Are electrolysis reactions thermodynamically spontaneous? Explain. (c) What process occurs at the anode in the electrolysis of molten NaCl? (d) Why is sodium metal not obtained when an aqueous solution of \(\mathrm{NaCl}\) undergoes electrolysis?

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

A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3},\) and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\). The overall cell reaction is $$2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s)$$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the \(\mathrm{Al}\) is not coated with its oxide.

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