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 \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?(\mathbf{c})\) What is the emf of the cell when \(\left[\mathrm{Ni}^{2+}\right]=0.200 M\) and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)

Short answer

(a) The emf of the cell under standard conditions is 0.513 V. (b) The emf of the cell when \([\mathrm{Ni}^{2+}]=3.00 \mathrm{M}\) and \([\mathrm{Zn}^{2+}]=0.100 \mathrm{M}\) is 0.557 V. (c) The emf of the cell when \([\mathrm{Ni}^{2+}]=0.200 M\) and \([\mathrm{Zn}^{2+}]=0.900 \mathrm{M}\) is 0.460 V.

Step-by-step solution

Find the standard reduction potentials

To determine the cell potential, first, look up the standard reduction potentials for both half-reactions: Zn^2+(aq) + 2e^− → Zn(s); \(E^⦵_{\text{red}}\) = −0.763 V Ni^2+(aq) + 2e^− → Ni(s); \(E^⦵_{\text{red}}\) = −0.250 V

Calculate the cell potential under standard conditions

Now, use the standard reduction potentials to determine the cell potential under standard conditions: \(E_{\text{cell}}^⦵\) = \(E_{\text{cathode}}^⦵\) - \(E_{\text{anode}}^⦵\) The cathode is where the reduction occurs (Ni^2+ is reduced to Ni), and the anode is where oxidation occurs (Zn is oxidized to Zn^2+). \(E_{\text{cell}}^⦵\) = (−0.250 V) - (−0.763 V) = 0.513 V So, the emf of the cell under standard conditions is 0.513 V.

Apply the Nernst Equation

To find the emf of the cell under non-standard conditions, we can apply the Nernst Equation: \(E_{\text{cell}}=E_{\text{cell}}^⦵-\frac{RT}{nF} \ln{Q}\) At 298 K, the simplified Nernst Equation is: \(E_{\text{cell}}=E_{\text{cell}}^⦵-\frac{0.0592}{n} \log{Q}\) In this case, n=2 since there are two electrons transferred in the reaction. Now, we can plug in the given concentrations for each of the conditions:

Cell potential with [Ni^2+]=3.00 M and [Zn^2+]=0.100 M

\(Q=\frac{[\text{Zn}^{2+}]}{[\text{Ni}^{2+}]}\) = \(\frac{0.100}{3.00}\) \(E_{\text{cell}}=0.513 - \frac{0.0592}{2}\log{\frac{0.100}{3.00}}\) = 0.513 + 0.0444 = 0.557 V The emf of the cell under these conditions is 0.557 V.

Cell potential with [Ni^2+]=0.200 M and [Zn^2+]=0.900 M

\(Q=\frac{[\text{Zn}^{2+}]}{[\text{Ni}^{2+}]}\) = \(\frac{0.900}{0.200}\) \(E_{\text{cell}}=0.513 - \frac{0.0592}{2}\log{\frac{0.900}{0.200}}\) = 0.513 - 0.0530 = 0.460 V The emf of the cell under these conditions is 0.460 V.

Key concepts

EMF Calculation

Electromotive force (emf) is a measure of the energy provided by a voltaic cell per charge that travels through the circuit. In simple terms, it's the voltage generated by the cell. To calculate the emf under standard conditions, we follow a specific process:

• Identify the chemical reaction: Determine the reaction occurring in the voltaic cell and what is oxidized and reduced.
• Use the formula for standard cell potential: The emf under standard conditions (\(E_{\text{cell}}^⦵\)) can be calculated using the formula:\[E_{\text{cell}}^⦵ = E_{\text{cathode}}^⦵ - E_{\text{anode}}^⦵\]Here, \(E_{\text{cathode}}^⦵\) stands for the standard reduction potential of the cathode, and \(E_{\text{anode}}^⦵\) is for the anode.
• Determine half-reactions: Find the half-reactions for the anode and cathode with their respective standard reduction potentials from a standard table.
• Subtract the anode potential from the cathode potential: This gives the emf under standard conditions. For example, in the given exercise, the calculated 📈 emf is \(0.513 \text{ V}\).

Understanding emf calculation is crucial for predicting how much work a cell can perform.

Standard Reduction Potentials

Standard reduction potentials are a fundamental concept in electrochemistry. They signify how readily a species gains or loses electrons. Standard conditions refer to solutions with 1 M concentration, gases at 1 atm pressure, and a temperature of 298 K.

• Definition: It's the voltage associated with a reduction reaction at an electrode when all substances are in their standard states.
• Significance: It helps determine which substance acts as the reducing or oxidizing agent. It indicates how easily a substance will undergo reduction compared to others.
• Using the values: Access standard reduction potential tables to find the required values. You will subtract these values one from another to calculate the overall cell potential.

In the given cell reaction, the standard reduction potential for zinc \(Zn^{2+} + 2e^- \rightarrow Zn\) is −0.763 V, whereas for nickel \(Ni^{2+} + 2e^- \rightarrow Ni\) is −0.250 V. This data is essential for calculating the standard emf of the cell. The more positive the potential, the greater the tendency to gain electrons, making it a stronger oxidizing agent.

Nernst Equation

The Nernst Equation is a vital tool in electrochemistry for evaluating the emf of a cell under non-standard conditions. It accounts for deviations from standard state conditions, like changes in concentration or temperature.

• Basics: The Nernst equation adjusts the standard emf by considering the concentration of reactants and products.
• Equation form: At room temperature (298 K), the simplified form is:\[E_{\text{cell}} = E_{\text{cell}}^⦵ - \frac{0.0592}{n} \log{Q}\]where \(n\) is the number of moles of electrons transferred in the reaction, and \(Q\) is the reaction quotient, calculated as \(\left[\text{products}\right]/\left[\text{reactants}\right]\)
• Calculation process: Substitute the concentrations into the equation to calculate the non-standard cell potential. For instance, when \([Ni^{2+}] = 3.00 \text{ M}\) and \([Zn^{2+}] = 0.100 \text{ M}\), \(Q\) equals \(\frac{0.100}{3.00}\), yielding an adjusted potential of \(0.557 \text{ V}\) for the cell.

By using the Nernst Equation, you can assess how concentration variances impact the ability of the cell to produce potential.