Problem 32 - Chemistry and Chemical Reactivity

Question

One half-cell in a voltaic cell is constructed from a copper wire electrode in a \(4.8 \times 10^{-3} \mathrm{M}\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\). The other half-cell consists of a zinc electrode in a \(0.40 \mathrm{M}\) solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\). Calculate the cell potential.

Step-by-Step Solution

Answer

The cell potential is approximately 1.0432 V.

Step 1: Identify Half-Reactions

The first step is to determine the reduction and oxidation half-reactions. For the copper half-cell:\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \] with a standard reduction potential \( E^0 = +0.34 \text{ V} \).For the zinc half-cell:\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \] with \( E^0 = -0.76 \text{ V} \).Since zinc is less noble, it will undergo oxidation. Thus its reverse reaction will be considered for oxidation.

Step 2: Write the Overall Cell Reaction

Combine the half-reactions with zinc as the oxidation half-cell and copper as the reduction half-cell. The overall reaction is:\[ \text{Zn} (s) + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu} (s) \]

Step 3: Determine Standard Cell Potential

Calculate the standard cell potential \( E^0_{cell} \) using:\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} \]Substitute the given values:\[ E^0_{cell} = 0.34 \text{ V} - (-0.76 \text{ V}) = 1.10 \text{ V} \]

Step 4: Use the Nernst Equation

Since the concentrations are not standard (1 M), use the Nernst equation to find the actual cell potential:\[ E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]Substitute \( n = 2 \), \( [\text{Zn}^{2+}] = 0.40 \text{ M} \), and \( [\text{Cu}^{2+}] = 4.8 \times 10^{-3} \text{ M} \):\[ E_{cell} = 1.10 \text{ V} - \frac{0.0592}{2} \log \left( \frac{0.40}{4.8 \times 10^{-3}} \right) \]

Step 5: Solve for the Cell Potential

First, calculate the reaction quotient:\[ \log \left( \frac{0.40}{4.8 \times 10^{-3}} \right) = \log (83.33) \approx 1.92 \]Solve for \( E_{cell} \):\[ E_{cell} = 1.10 \text{ V} - 0.0296 \times 1.92 \]\[ E_{cell} = 1.10 \text{ V} - 0.0568 \text{ V} = 1.0432 \text{ V} \]

Step 6: State the Final Result

The calculated cell potential for the voltaic cell is approximately \( E_{cell} = 1.0432 \text{ V} \).

Key Concepts

Voltaic Cells

Voltaic cells, also known as galvanic cells, are devices that convert chemical energy into electrical energy. They consist of two different electrodes placed in separate electrolyte solutions and connected by a conductive pathway. Each electrode is dipped into an electrolyte that contains ions which can undergo oxidation or reduction.

In a voltaic cell, one electrode functions as the anode and the other as the cathode. The anode is the site of oxidation, where electrons are released, while the cathode is where reduction occurs, accepting electrons. This movement of electrons through the external circuit creates an electric current, which can be harnessed to perform work.

The function of a voltaic cell is reminiscent of a battery that powers our electronic devices. By combining specific metal electrodes in an electrolyte, electricity is generated spontaneously. The copper and zinc electrodes from the original exercise are classic examples of materials used in such cells. Copper acts as the cathode, while zinc serves as the anode, setting up an operational cell when connected by a wire.

Nernst Equation

The Nernst equation is a crucial tool in electrochemistry to determine the actual potential of an electrochemical cell, taking into account deviations from standard conditions. Standard conditions assume all reactants and products are at 1 M concentration. However, that's often not the case in real-world applications. The Nernst equation bridges this gap.

It is mathematically expressed as: \[ E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log \left( \frac{\text{concentration of products}}{\text{concentration of reactants}} \right) \] Here,

• \(E_{cell}\) is the cell potential under non-standard conditions,
• \(E^0_{cell}\) is the standard cell potential,
• \(n\) is the number of moles of electrons transferred in the reaction,
• \(\text{concentration of products/reactants}\) refers to their molarity.

This equation shows that the cell potential changes depending on the concentrations of the ions in solution. By using the Nernst equation, one can predict how the cell voltage will vary as the reaction progresses or as the concentrations of reactants and products change.

Cell Potential Calculation

Cell potential calculation involves determining the voltage generated by a voltaic cell. For standard conditions, it is straightforward: simply subtract the anode's reduction potential from the cathode's. In the example given, copper serves as the cathode with a standard reduction potential of \(0.34\, \text{V}\) and zinc as the anode with \(-0.76\, \text{V}\). This results in a standard cell potential of: \[ E^0_{cell} = 0.34 \text{ V} - (-0.76 \text{ V}) = 1.10 \text{ V} \] However, since the concentrations differ from the standard 1 M, the Nernst equation is employed to adjust this value. For the calculation:

• Identify the concentration of ions involved: \([\text{Cu}^{2+}] = 4.8 \times 10^{-3} \text{ M}\) and \([\text{Zn}^{2+}] = 0.40 \text{ M}\).
• Substitute these into the Nernst equation, using \(n = 2\) as the moles of electrons exchanged.
• Compute the reaction quotient log term and subsequently, the corrected cell potential.

After this adjustment, cell potential comes out to approximately \(1.0432\, \text{V}\). This reflects the real-world potential generated by the cell under the given conditions.