Problem 97 - Chemistry The Molecular Science

Question

The permanganate ion \(\mathrm{MnO}_{4}^{-}\) can be reduced to the manganese(II) ion \(\mathrm{Mn}^{2+}\) in aqueous acidic solution, and the half- cell potential for this half-cell reaction is \(1.51 \mathrm{~V}\). If this half-cell is combined with a \(\mathrm{Zn}^{2+} \mid \mathrm{Zn}\) half-cell to form a voltaic cell at standard conditions, (a) Write the chemical equation for the half-reaction occurring at the anode. (b) Write the chemical equation for the half-reaction occurring at the cathode. (c) Write the overall balanced equation for the reaction. (d) Calculate the cell potential.

Step-by-Step Solution

Answer

The cell potential is 2.27 V.

Step 1: Identify Anode and Cathode

In a voltaic cell, the anode is where oxidation occurs, and the cathode is where reduction occurs. Given the half-cells are permanganate ion reduction and zinc oxidation, the Zn/Zn²⁺ half-cell acts as the anode, and the MnO₄⁻ reduction half-cell acts as the cathode.

Step 2: Write the Anode Half-Reaction

At the anode, oxidation of zinc occurs. The half-reaction is: \[ \mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2e^- \] This represents the loss of electrons by zinc as it forms zinc ions.

Step 3: Write the Cathode Half-Reaction

At the cathode, the reduction of permanganate ions takes place. The half-reaction is: \[ \mathrm{MnO}_4^-(aq) + 8\mathrm{H}^+(aq) + 5e^- \rightarrow \mathrm{Mn}^{2+}(aq) + 4\mathrm{H}_2\mathrm{O}(l) \] This represents the gain of electrons by permanganate ions in acidic solution.

Step 4: Balance the Overall Reaction

To balance the overall reaction, combine the two half-reactions and balance electron transfer. Multiply the zinc reaction by 5, and the permanganate reaction by 2, to equalize electron transfer: \[5\mathrm{Zn}(s) + 2\mathrm{MnO}_4^-(aq) + 16\mathrm{H}^+(aq) \rightarrow 5\mathrm{Zn}^{2+}(aq) + 2\mathrm{Mn}^{2+}(aq) + 8\mathrm{H}_2\mathrm{O}(l) \]This equation is the balanced overall cell reaction.

Step 5: Calculate the Cell Potential

The standard cell potential is calculated using the standard reduction potentials: \[ E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ \]Given that the potential for MnO₄⁻ reduction is 1.51 V, and the standard reduction potential for Zn²⁺/Zn is -0.76 V, \[ E_{\text{cell}}^\circ = 1.51 \mathrm{~V} - (-0.76 \mathrm{~V}) = 2.27 \mathrm{~V} \]

Key Concepts

Voltaic Cell

A voltaic cell, also known as a galvanic cell, is a device that harnesses the chemical energy from spontaneous redox reactions to generate electrical energy. It consists of two half-cells, each containing an electrode and an electrolyte solution. The electrolyte contains ions that can participate in the redox reactions.
In a voltaic cell, oxidation occurs at the anode and reduction occurs at the cathode. Electrons flow from the anode to the cathode through an external circuit. The anode is negatively charged, while the cathode is positively charged.

Voltaic cells are crucial for many everyday applications, such as batteries. In essence, they convert chemical energy into electricity, which is a reversible process in rechargeable batteries. Using the example in the original exercise, the zinc half-cell acts as the anode, where zinc metal is oxidized to zinc ions. The permanganate reduction half-cell is the cathode, where electrons are gained and permanganate ions are reduced.

Reduction Reaction

A reduction reaction is a part of a redox (reduction-oxidation) process where a substance gains electrons. This gain of electrons results in a decrease in the oxidation state of the substance. Reduction reactions are integral to many chemical and biological processes.

In the context of a voltaic cell, the reduction reaction takes place at the cathode. Here, a species with a higher electron affinity gains electrons. This is known as a reduction reaction because it "reduces" the charge of the ion involved. Electrons flow towards this half of the voltaic cell because of its higher electric potential.
In the given exercise, the permanganate ion (\(\text{MnO}_4^-\)) is reduced to manganese (II) ion (\(\text{Mn}^{2+}\)) in an acidic solution. The balanced half-reaction is:

• \[\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)\]

This reaction highlights the transformation of MnO\(^-_4\) as it gains electrons and protons to form \(\text{Mn}^{2+}\).

Standard Cell Potential

The standard cell potential (\(E_{\text{cell}}^\circ\)) is a measure of the voltage difference between two half-cells under standard conditions, which is typically 1 M concentration, 1 atm pressure, and 25°C (298 K) temperature. It indicates the maximum potential difference and the capability of the cell to do work.
Standard cell potential is calculated by taking the difference between the standard reduction potentials of the cathode and the anode:

• \[E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ\]

In the original exercise, the reduction potential for MnO\(_4^-\) is given as 1.51 V, and the reduction potential for the Zn\(^{2+}\)/Zn couple is -0.76 V.
By substituting these values into the formula, we obtain the standard cell potential:

• \[E_{\text{cell}}^\circ = 1.51 \text{ V} - (-0.76 \text{ V}) = 2.27 \text{ V}\]

This positive value indicates a spontaneous process, meaning the reactions can occur naturally and the cell can generate an electric current.