Question

What is the standard cell potential and the net reaction in a galvanic cell that has the following half reactions? $$ \begin{aligned} \mathrm{MnO}_{2}+4 \mathrm{H}^{+}+2 e^{-} \rightleftharpoons \mathrm{Mn}^{2+} &+2 \mathrm{H}_{2} \mathrm{O} \\ & E_{\mathrm{MnO}_{2}}^{\circ}=1.23 \mathrm{~V} \\ \mathrm{PbCl}_{2}+2 e^{-} \rightleftharpoons \mathrm{Pb}+2 \mathrm{Cl}^{-} & E_{\mathrm{PbCl}}^{\mathrm{o}}=-0.27 \mathrm{~V} \end{aligned} $$

Short answer

The standard cell potential of the galvanic cell is 1.50 V, and the net reaction is \(\mathrm{MnO}_{2} + \mathrm{PbCl}_{2} + 4 \mathrm{H}^{+} \rightarrow \mathrm{Pb} + \mathrm{Mn}^{2+} + 2 \mathrm{Cl}^{-} + 2 \mathrm{H}_2\mathrm{O}\).

Step-by-step solution

Identify the oxidation and reduction half-reactions

First, identify which half-reaction is oxidation and which is reduction. The oxidation reaction is the one where electrons are produced (increase in oxidation state), while the reduction reaction is the one where electrons are consumed (decrease in oxidation state). The given half-reactions are already written as reductions, so we need to determine which should be reversed to act as the oxidation in the galvanic cell.

Determine the direction of the half-reactions

The cell potential is positive when the reduction potential for the cathode (reduction) is greater than the anode (oxidation). Since \(E_{\mathrm{MnO}_2}^\circ = 1.23 \mathrm{~V}\) is greater than \(E_{\mathrm{PbCl}_2}^\circ = -0.27 \mathrm{~V}\), the half-reaction with \(\mathrm{MnO}_2\) will act as the reduction half-reaction at the cathode, and the half-reaction with \(\mathrm{PbCl}_2\) will be reversed to act as the oxidation half-reaction at the anode.

Calculate the standard cell potential

The standard cell potential \(E^\circ_{\text{cell}}\) is obtained by subtracting the standard reduction potential of the anode \(E^o_{\text{anode}}\) from the standard reduction potential of the cathode \(E^\circ_{\text{cathode}}\). In this case, since the \(\mathrm{PbCl}_2\) reaction is the anode and is being reversed, we change the sign of its standard potential before subtracting: \[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.23 \mathrm{V} - (-0.27 \mathrm{V}) = 1.50 \mathrm{V}\].

Write the net reaction

Combine the two half-reactions to get the net reaction. Remember to reverse the anode reaction and balance the number of electrons. The electrons should cancel out in the net equation:\[\mathrm{PbCl}_{2} + 2 e^{-} \rightarrow \mathrm{Pb} + 2 \mathrm{Cl}^{-} \text{ (Reversed)}\]\[\mathrm{PbCl}_{2} \rightarrow \mathrm{Pb} + 2 \mathrm{Cl}^{-} + 2 e^{-}\]\[\mathrm{MnO}_{2} + 4 \mathrm{H}^{+} + 2 e^{-} \rightarrow \mathrm{Mn}^{2+} + 2 \mathrm{H}_{2}\mathrm{O}\]Combine them to get the net reaction: \[\mathrm{MnO}_{2} + \mathrm{PbCl}_{2} + 4 \mathrm{H}^{+} \rightarrow \mathrm{Pb} + \mathrm{Mn}^{2+} + 2 \mathrm{Cl}^{-} + 2 \mathrm{H}_2\mathrm{O}\]

Key concepts

Galvanic Cell

A Galvanic cell, also known as a voltaic cell, is a device that converts chemical energy into electrical energy through a spontaneous redox reaction. It consists of two half-cells connected by a salt bridge or porous membrane that allows ions to move between them, while preventing the mixing of different solutions that may lead to precipitation or other side reactions.

In a galvanic cell, each half-cell contains an electrode and an electrolyte. One half-cell acts as the anode where oxidation occurs, and the other acts as the cathode where reduction takes place. Electrons flow from the anode to the cathode through an external circuit, which is harnessed to do work, like powering a light bulb. The movement of ions in the cell completes the electrical circuit by maintaining charge balance.

Half-Reactions

In electrochemistry, half-reactions are at the heart of analyzing galvanic cells. They show the two parts of a redox reaction: oxidation and reduction. Oxidation half-reactions detail the loss of electrons by a species, while reduction half-reactions show the gain of electrons. In a balanced redox reaction, the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

For clarity and easier understanding, identifying each half-reaction is crucial. It allows you to see which species are being oxidized and reduced, and to ensure that electron transfer between the two is balanced. This is an essential step in solving galvanic cell problems and determining the overall cell potential.

Oxidation and Reduction

The concepts of oxidation and reduction are foundational to understanding galvanic cells. Oxidation refers to the loss of electrons by a substance, indicated by an increase in oxidation state. Conversely, reduction is the gain of electrons, which reduces the oxidation state of the substance. These two processes occur simultaneously in a redox reaction; as one species loses electrons (oxidized), another gains those electrons (reduced).

To determine the direction of the half-reactions in a galvanic cell, you analyze their standard potentials. The species with the higher reduction potential typically undergoes reduction at the cathode, while the other, with a lower or negative potential, undergoes oxidation at the anode. Remembering the mnemonic 'OIL RIG' (Oxidation Is Loss, Reduction Is Gain) can help recall which process involves the loss or gain of electrons.

Net Reaction

The net reaction in a galvanic cell is the sum of the oxidation and reduction half-reactions. Once the directions of the half-reactions are established, they are combined, ensuring that the electrons cancel out. This final balanced equation represents the total chemical change occurring within the galvanic cell.

When writing the net reaction, it's vital to reverse the oxidation half-reaction if it is initially given in the reduction format, as in the textbook solution example. And lastly, balancing the overall reaction for mass and charge is imperative for an accurate representation of the galvanic cell's operation. Understanding the net reaction provides insight into the chemical processes at play, and allows for the computation of the cell's theoretical voltage output or standard cell potential.