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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) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel-metal-hydride batteries over nickel-cadmium batteries?

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

A \(1 M\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a beaker with a strip of Cu metal. A \(1 M\) solution of \(\mathrm{SnSO}_{4}\) is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode and which as the cathode? (b) Which electrode gains mass and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger oxidizing agent: (a) \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\mathrm{Cd}^{2+}(a q)\) (c) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{3}(g)\)

Gold exists in two common positive oxidation states, +1 and +3. The standard reduction potentials for these oxidation states are $$ \begin{aligned} \mathrm{Au}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) & E_{\mathrm{red}}^{o}=+1.69 \mathrm{~V} \\ \mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) & E_{\mathrm{red}}^{\circ}=+1.50 \mathrm{~V} \end{aligned} $$ (a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{aligned} 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+& 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \\ & 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) \end{aligned} $$ What is being oxidized and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with Zn dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced?

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