Question

Assume the specifications of a Ni-Cd voltaic cell include delivery of 0.25 A of current for \(1.00 \mathrm{h}\). What is the minimum mass of the cadmium that must be used to make the anode in this cell?

Short answer

The minimum mass of cadmium needed is 0.524 grams.

Step-by-step solution

Calculate Total Charge

First, we need to calculate the total charge that flows through the voltaic cell. Given that current, \( I = 0.25 \) A, and time, \( t = 1 \text{ hour} = 3600 \) seconds, we can calculate charge \( Q \) using the formula \( Q = I \times t \). Thus, \( Q = 0.25 \times 3600 = 900 \) coulombs.

Apply Faraday's Law

Next, we use Faraday's law to relate charge to the amount of substance involved in the electrochemical reaction. The charge on one mole of electrons (Faraday's constant) is \( 96485 \) coulombs. The reaction at the anode is \( ext{Cd} ightarrow ext{Cd}^{2+} + 2 ext{e}^{-} \). This reaction shows that 2 moles of electrons are needed per mole of cadmium, which means \( 96500 \times 2 = 193000 \) coulombs are needed per mole of cadmium.

Calculate Moles of Cadmium

Now, calculate the moles of cadmium that correspond to the total charge. Use the formula \( ext{moles of Cd} = \frac{Q}{2 \times 96485} \). Substituting the values gives \( ext{moles of Cd} = \frac{900}{193000} = 0.00466 \) moles.

Convert Moles to Mass

Finally, convert the moles of cadmium to mass. The molar mass of cadmium is \( 112.41 \text{ g/mol} \). Therefore, the mass of cadmium needed is \( ext{mass} = ext{moles} \times ext{molar mass} = 0.00466 \times 112.41 = 0.524 \text{ grams} \).

Key concepts

Voltaic Cells

A voltaic cell, also known as a galvanic cell, is an electrochemical cell that generates electricity through spontaneous redox reactions. These cells harness the energy released from chemical reactions to produce an electric current. In a voltaic cell, two different metals are placed in separate compartments, each, known as half-cells.
Here's a simple breakdown of how a voltaic cell works:

• Each half-cell contains a metal electrode submerged in a solution containing ions of the same metal.
• The metal that loses electrons undergoes oxidation and forms ions, acting as the anode.
• The metal that gains electrons undergoes reduction and acts as the cathode.
• The electron flow between these electrodes generates electricity, which can be harnessed to perform work.

Understanding this basic setup helps us see how the Ni-Cd cell in the original problem functions to provide electrical energy over time.

Faraday's Law

Faraday's Law is essential in electrochemistry, linking the amount of electric charge with the amount of chemical change. It states that the amount of substance dissolved or deposited at an electrode during electrolysis is proportional to the total electric charge passed through the substance.
Key points to understand about Faraday's Law:

• 1 mole of electrons has a charge of 96,485 coulombs, known as Faraday's constant.
• In the context of the question, the anodic reaction, where Cd is oxidized to Cd²⁺, makes use of this law. It shows that 2 moles of electrons are needed for each mole of cadmium oxidized.
• The relationship can be expressed with the formula: \[ Q = n_e \times 96,485 \], where \( n_e \) is the moles of electrons.

By applying this law, we were able to determine the precise amount of cadmium involved in the electrochemical process, helping to find out the mass needed.

Moles to Mass Conversion

Converting moles to mass is a fundamental concept in chemistry, allowing us to determine how much of a substance is present in terms of mass, given the amount in moles. This conversion requires knowledge of the substance's molar mass or molecular weight.
To calculate mass from moles:

• Use the equation: \[ \text{mass} = \text{moles} \times \text{molar mass} \]
• The molar mass of cadmium, from the periodic table, is 112.41 g/mol.
• In the problem, with 0.00466 moles of cadmium determined through previous calculations, we use this formula to find the mass: \[ \text{mass} = 0.00466 \times 112.41 = 0.524 \text{grams} \]

Mastering moles to mass conversion enhances understanding of quantitative chemical computations, and is a crucial step in solving many electrochemistry problems.