Problem 21 - Chemistry The Science in Context

Question

An electrochemical cell with an aqueous electrolyte is based on the reaction between \(\mathrm{Ni}^{2+}(a q)\) and \(\mathrm{Cd}(s)\) producing \(\mathrm{Ni}(s)\) and \(\mathrm{Cd}^{2+}(a q)\) a. Write half-reactions for the anode and cathode. b. Write a balanced net ionic equation describing the cell reaction. c. Draw the cell diagram.

Step-by-Step Solution

Answer

Reduction half-reaction (Cathode): \(\mathrm{Ni}^{2+}(aq) + 2e^− → \mathrm{Ni}(s)\) Oxidation half-reaction (Anode): \(\mathrm{Cd}(s) → \mathrm{Cd}^{2+}(aq) + 2e^−\) 2. What is the balanced net ionic equation for the cell reaction? \(\mathrm{Ni}^{2+}(aq) + \mathrm{Cd}(s) → \mathrm{Ni}(s) + \mathrm{Cd}^{2+}(aq)\) 3. What is the cell diagram of the electrochemical cell? \(\mathrm{Cd(s)|Cd^{2+}(aq)||Ni^{2+}(aq)|Ni(s)}\)

Step 1: Identify the oxidation and reduction half-reactions

First, we'll read the given information to find which species undergo oxidation and reduction. The reaction converts \(\mathrm{Ni}^{2+}(aq)\) to \(\mathrm{Ni}(s)\) and \(\mathrm{Cd}(s)\) to \(\mathrm{Cd}^{2+}(aq)\). Since \(\mathrm{Ni}^{2+}(aq)\) gains electrons and becomes \(\mathrm{Ni}(s)\), it undergoes reduction. On the other hand, \(\mathrm{Cd}(s)\) loses electrons and becomes \(\mathrm{Cd}^{2+}(aq)\), so it undergoes oxidation.

Step 2: Write the balanced half-reactions

Write the half-reactions for each process, making sure that they are balanced with respect to atoms and charge: Reduction (Cathode): \(\mathrm{Ni}^{2+}(aq) + 2e^− → \mathrm{Ni}(s)\) Oxidation (Anode): \(\mathrm{Cd}(s) → \mathrm{Cd}^{2+}(aq) + 2e^−\)

Step 3: Write the balanced net ionic equation for the cell reaction

To find the balanced net ionic equation, we can combine the balanced half-reactions: \(\mathrm{Ni}^{2+}(aq) + 2e^− → \mathrm{Ni}(s)\) + \(\mathrm{Cd}(s) → \mathrm{Cd}^{2+}(aq) + 2e^−\) The electrons cancel out, and we get: \(\mathrm{Ni}^{2+}(aq) + \mathrm{Cd}(s) → \mathrm{Ni}(s) + \mathrm{Cd}^{2+}(aq)\)

Step 4: Draw the cell diagram

To draw the cell diagram, follow these steps: 1. Identify the anode, where oxidation takes place, and the cathode, where reduction occurs. 2. Write the half-reactions as shown at each electrode. 3. Use a double vertical line (||) to represent the salt bridge that allows the flow of ions between the two half-cells. 4. Use a single vertical line (|) to separate the elements and ions in each half-cell. The cell diagram for this electrochemical cell is: \(\mathrm{Cd(s)|Cd^{2+}(aq)||Ni^{2+}(aq)|Ni(s)}\)

Key Concepts

Anode and Cathode Half-Reactions

To understand an electrochemical cell's function, distinguishing between anode and cathode half-reactions is fundamental. The anode is where oxidation takes place; it's the spot that gives up electrons. In contrast, the cathode is where reduction occurs, meaning it's where the electrons are accepted. In the exercise presented, Oxidation (Anode):
\(\mathrm{Cd(s) \rightarrow Cd^{2+}(aq) + 2e^−}\)
represents the half-reaction happening at the anode, with solid cadmium (\(\mathrm{Cd(s)}\)) transforming into cadmium ions and releasing two electrons in the process. On the other side, we have:
Reduction (Cathode):
\(\mathrm{Ni^{2+}(aq) + 2e^− \rightarrow Ni(s)}\)
which describes the reaction at the cathode where nickel ions (\(\mathrm{Ni^{2+}(aq)}\)) gain electrons to become solid nickel (\(\mathrm{Ni(s)}\)). These half-reactions are crucial for balancing the overall reaction and ensuring the conservation of charge.

Balanced Net Ionic Equation

The balanced net ionic equation is the result of combining the anode and cathode half-reactions, subtracting anything that appears on both sides of the equation. Effectively, it represents the core chemical change in an electrochemical cell. After cancellation of electrons, for this cell, we arrive at the following succinct equation: \(\mathrm{Ni^{2+}(aq) + Cd(s) \rightarrow Ni(s) + Cd^{2+}(aq)}\)
This equation tells us what is empirically happening: nickel ions in the aqueous phase are swapping places with solid cadmium, generating solid nickel and cadmium ions in solution. While balancing these equations, ensure that both mass and charge are conserved; the number of atoms of each element and the total charge must be the same on both sides of the reaction.

Cell Diagram

The cell diagram visualizes the setup of an electrochemical cell. In simple terms, it's a shorthand notation to represent the arrangement of components. The example from the exercise can be construed as:
\(\mathrm{Cd(s)|Cd^{2+}(aq)||Ni^{2+}(aq)|Ni(s)}\)
Here is how to read it: Cadmium metal (\(\mathrm{Cd(s)}\)) at the anode is in contact with its own ions in solution (\(\mathrm{Cd^{2+}(aq)}\)), divided from the cathode half where aqueous nickel ions (\(\mathrm{Ni^{2+}(aq)}\)) are in contact with solid nickel (\(\mathrm{Ni(s)}\)). The salt bridge is represented by the double vertical line '||', with single vertical lines '|' used to separate different states of matter within each half-cell. It is important that the cell diagram is arranged correctly to reflect the direction of electron flow and ionic movements within the cell.

Oxidation and Reduction

Grasping the concepts of oxidation and reduction is central to understanding electrochemistry. In its simplest terms, oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Remember the mnemonic 'OIL RIG': Oxidation Is Loss, Reduction Is Gain. Looking at our exercise, cadmium undergoes oxidation, losing two electrons:
\(\mathrm{Cd(s) \rightarrow Cd^{2+}(aq) + 2e^−}\)
Meanwhile, nickel is reduced as it gains electrons:
\(\mathrm{Ni^{2+}(aq) + 2e^− \rightarrow Ni(s)}\)
These processes always occur simultaneously, with the electrons lost by the oxidized species being gained by the reduced species, which is a fundamental concept called the conservation of charge. To capture the full essence of electrochemical cells, one must become confident in distinguishing between these two complementary processes.