Question

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 \(\left[Z n^{2+}\right]=0.100 \mathrm{M}\) ? (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

(a) The emf of the cell under standard conditions is \(0.53 \mathrm{V}\). (b) The emf of the cell when \([\mathrm{Ni}^{2+}]=3.00 \mathrm{M}\) and \([\mathrm{Zn}^{2+}]=0.100 \mathrm{M}\) is \(0.5596 \mathrm{V}\). (c) The emf of the cell when \([\mathrm{Ni}^{2+}]=0.200 \mathrm{M}\) and \([\mathrm{Zn}^{2+}]=0.900 \mathrm{M}\) is \(0.5004 \mathrm{V}\).

Step-by-step solution

Finding the emf under standard conditions.

First, we need to look up the standard reduction potentials for the involved half-reactions in a textbook or reference table. The half-reactions are: Zn²⁺(aq) + 2e⁻ → Zn(s) E⁰ (Zn²⁺/Zn) = -0.76 V Ni²⁺(aq) + 2e⁻ → Ni(s) E⁰ (Ni²⁺/Ni) = -0.23 V Since Zn is losing electrons and Ni²⁺ is gaining them, the Zn half-reaction is the anode (oxidation) and the Ni half-reaction is the cathode (reduction). To get the overall emf of the cell (E⁰_cell), we subtract the anode potential from the cathode potential: E⁰_cell = E⁰_cathode - E⁰_anode

Calculate the emf under standard conditions.

Now, let's calculate the emf of the cell using the standard reduction potentials from step 1: E⁰_cell = (-0.23 V) - (-0.76 V) = 0.53 V So, under standard conditions, the emf of the cell is 0.53 V.

Apply Nernst equation for different concentrations.

We can now use the Nernst equation to find the emf of the cell given different concentrations: E_cell = E⁰_cell - \(\frac{RT}{nF}\) ln(Q) Where, E_cell: emf of the cell R: Gas constant (8.314 J/(K·mol)) T: Temperature (298 K) n: number of electrons transferred (2 in this case) F: Faraday's constant (96485 C/mol) Q: Reaction quotient, given by \(\frac{[Zn^{2+}]}{[Ni^{2+}]}\)

Calculate the emf for the given concentrations.

We can now calculate the emf for the given concentrations of Ni²⁺ and Zn²⁺ ions: (b) [Ni²⁺] = 3.00 M, and [Zn²⁺] = 0.100 M Q = \(\frac{0.100}{3.00}\) = 0.0333 E_cell = 0.53 V - \(\frac{8.314 \times 298}{2 \times 96485}\) ln(0.0333) = 0.53 V + 0.0296 V = 0.5596 V So, the emf of the cell when [Ni²⁺] = 3.00 M and [Zn²⁺] = 0.100 M is 0.5596 V. (c) [Ni²⁺] = 0.200 M, and [Zn²⁺] = 0.900 M Q = \(\frac{0.900}{0.200}\) = 4.50 E_cell = 0.53 V - \(\frac{8.314 \times 298}{2 \times 96485}\) ln(4.50) = 0.53 V - 0.0296 V = 0.5004 V So, the emf of the cell when [Ni²⁺] = 0.200 M and [Zn²⁺] = 0.900 M is 0.5004 V.

Key concepts

Understanding Voltaic Cells

Voltaic cells, also known as galvanic cells, are devices that convert chemical energy into electrical energy through spontaneous redox reactions. In a voltaic cell, two different metals are placed in solutions of their respective ions, forming two half-cells.
Each half-cell contains a metal electrode and an electrolyte solution.
One half-cell undergoes oxidation (anode) and the other reduction (cathode). Key Points:

• Anode: The site of oxidation where electrons are released.
• Cathode: The site of reduction where electrons are gained.
• Salt Bridge: Maintains charge balance by allowing the flow of ions.
• Electromotive Force (emf): The voltage or potential difference created by the chemical reaction.

Understanding these components helps us analyze how cells produce electricity and how emf is calculated under different conditions.

Exploring the Nernst Equation

The Nernst Equation is a vital tool in electrochemistry, allowing us to calculate the cell potential under non-standard conditions. It adjusts the standard cell potential to consider the effect of ion concentrations.The equation is:\[E_{cell} = E^°_{cell} - \frac{RT}{nF} \ln(Q)\]Where:

• \(E_{cell}\): Cell potential at non-standard conditions.
• \(E^°_{cell}\): Standard cell potential.
• \(R\): Universal gas constant \((8.314 \text{ J/(K·mol)})\).
• \(T\): Temperature in Kelvin.
• \(n\): Number of electrons transferred in the reaction.
• \(F\): Faraday's constant \((96485 \text{ C/mol})\).
• \(Q\): Reaction quotient.

This equation is crucial in determining how changes in concentration affect the emf of a voltaic cell.

Standard Reduction Potential

The standard reduction potential is a measure of the tendency of a chemical species to acquire electrons and be reduced. These potentials are measured under standard conditions: \(1\) M concentration, \(1\) atm pressure, and \(298 \text{ K}\).Important Details:

• Measured in volts (V).
• Higher positive value indicates a greater tendency to gain electrons.
• Essential in calculating the overall cell potential by subtracting the anode potential from the cathode potential:

\[E^°_{cell} = E^°_{cathode} - E^°_{anode}\]Using these values, we determine the direction and feasibility of the reaction, helping us understand which half-reaction occurs at the cathode and which at the anode.

The Role of the Reaction Quotient

The reaction quotient \(Q\) is a dimensionless number that compares the relative amounts of products and reactants present during a reaction at a given point. It is calculated similarly to the equilibrium constant, but for any point in time, not just equilibrium.Formula:\[Q = \frac{[Zn^{2+}]}{[Ni^{2+}]}\]In a voltaic cell, \(Q\) reflects the concentration of ions involved in the redox reaction. It directly impacts the cell's emf when applied to the Nernst Equation:

• If \(Q < 1\), the forward reaction is favored, increasing the emf.
• If \(Q > 1\), the reverse reaction is favored, decreasing the emf.
• If \(Q = 1\), the system is at standard conditions, and \(E_{cell} = E^°_{cell}\).

Understanding \(Q\)'s impact is essential for predicting how changes in ion concentration affect the electrochemical cell's behavior.