Problem 53 - Chemistry The Science in Context

Question

An electrochemical cell consists of a standard hydrogen electrode and a second half-cell in which a magnesium electrode is immersed in a \(1.00 M\) solution of \(\mathrm{Mg}^{2+}\) ions. a. What is the value of \(E_{\text {cell }} ?\) b. Which electrode is the anode? c. Which is a product of the cell reaction: \(\mathrm{H}^{+}\) ions or \(\mathrm{H}_{2}\) gas?

Step-by-Step Solution

Answer

Answer: The electrochemical cell potential is 2.37 V. b. Which electrode is the anode? Answer: The anode is the magnesium electrode. c. Are \(\mathrm{H}^{+}\) ions or \(\mathrm{H}_2\) gas a product of the cell reaction? Answer: \(\mathrm{H}_2\) gas is a product of the cell reaction.

Step 1: Calculate the E0 values for both half-cells

Look up the standard reduction potentials (E0) for both half-cells involved in the electrochemical cell. For the standard hydrogen electrode (SHE): \(\mathrm{2H^{+} + 2e^{-}} \rightarrow \mathrm{H_{2} (g)}\) \(E_{\mathrm{SHE}}^0 = 0 \, V\) For the magnesium half-cell: \(\mathrm{Mg^{2+} + 2e^{-}} \rightarrow \mathrm{Mg (s)}\) We need to look up its standard reduction potential value in a table, and we find that \(E_{\mathrm{Mg}}^0 = -2.37 \, V\)

Step 2: Determine which half-cell will be reduced and which one will be oxidized

We need to see which half-cell is more likely to be reduced (gain electrons) based on the standard reduction potential. A more positive E0 value represents a greater tendency to be reduced. The standard hydrogen electrode has a higher E0 value (\(0 \, V\)) than the magnesium half-cell (\(-2.37 \, V\)). This means that the hydrogen half-cell will be reduced, and the magnesium half-cell will be oxidized.

Step 3: Calculate the Ecell

To calculate the \(E_{\text{cell}}\), use the formula: \(E_{\text{cell}} = E_{\text{cathode}}^0 - E_{\text{anode}}^0\) Since the hydrogen half-cell is being reduced, it is the cathode, and the magnesium half-cell is the anode (where oxidation occurs): \(E_{\text{cell}} = E_{\mathrm{SHE}}^0 - E_{\mathrm{Mg}}^0 = 0 \, V - (-2.37 \, V) = 2.37 \, V\) So the value of \(E_{\text{cell}} = 2.37 \, V\)

Step 4: Answer a, b, and c

a. The value of \(E_{\text{cell}}\) is \(2.37 \, V\). b. The anode is the magnesium electrode, where oxidation occurs. c. Since the hydrogen electrode is the cathode (where reduction occurs), it means that \(\mathrm{H}^{+}\) ions are being converted into \(\mathrm{H}_2\) gas as a product of the cell reaction.

Key Concepts

Understanding Standard Reduction Potential

An electrochemical cell works through the principle of redox reactions, where one substance is reduced while another is oxidized. The standard reduction potential, expressed as \( E^0 \), is a measure of the tendency of a chemical species to acquire electrons and be reduced. It is essential to understand how these potentials interplay in an electrochemical cell.

For reference, the standard electrode potential is determined under standard conditions, which includes a concentration of \(1 \, M\), a pressure of \(1 \, atm\), and a temperature of \(25 \, ^{\circ}C\). Lists of standard reduction potentials help to predict the direction of electron flow between the two half-cells in an electrochemical cell configuration.

In our scenario, we compare two potentials: the standard hydrogen electrode (SHE) with a potential of \(0 \, V\) and the magnesium electrode with a potential of \(-2.37 \, V\). The SHE has the higher potential, indicating it has a greater likelihood of being reduced than the magnesium half-cell. By knowing the \( E^0 \) values, we can determine the roles of anode and cathode, essential steps in building an electrochemical cell.

Exploring Half-Cell Reactions

Half-cell reactions form the building blocks of an electrochemical cell. Each half-cell consists of a metal electrode submerged in a solution of its ions, and the reaction here can either involve oxidation or reduction. This is depicted through the process of gaining or losing electrons.

In our example, we have two half-cells: the hydrogen half-cell where the reaction involves reduction:

• \( \mathrm{2H^{+} + 2e^{-}} \rightarrow \mathrm{H_{2} (g)} \)

And the magnesium half-cell where oxidation occurs:

• \( \mathrm{Mg (s)} \rightarrow \mathrm{Mg^{2+} + 2e^{-}} \)

By identifying which half-cell undergoes reduction or oxidation, we determine the roles within the cell: the cathode is where reduction happens, and the anode is where oxidation occurs. Thus, in this case, the magnesium acts as the anode, and hydrogen as the cathode. Understanding these reactions is crucial, as they dictate the flow of electrons and consequently the electrical energy produced.

Identifying Cell Reaction Products

When analyzing an electrochemical cell, it is vital to recognize the products of the cell reaction. These products are the result of the redox processes occurring at the cathode and anode.

In this problem, we need to determine whether \(\mathrm{H}^+\) ions or \(\mathrm{H}_2\) gas appears as a product. At the cathode, where the SHE is operating, reduction takes place, converting \(\mathrm{H}^+\) ions into \(\mathrm{H}_2\) gas:

\(\mathrm{2H^{+} + 2e^{-}} \rightarrow \mathrm{H_{2} (g)}\)

Thus, \(\mathrm{H}_2\) gas is the product of the cell reaction at the cathode. By identifying this, we can comprehensively understand both the conversion of ions into gases within the electrochemical cell and the overall energy conversion process that provides the electrical energy needed for various applications.