Problem 90 - Chemistry and Chemical Reactivity

Question

A You want to use electrolysis to plate a cylindrical object (radius \(=2.50 \text { and length }=20.00 \mathrm{cm})\) with a coating of nickel metal, \(4.0 \mathrm{mm}\) thick. You place the object in a bath containing a salt \(\left(\mathrm{Na}_{2} \mathrm{SO}_{4}\right) .\) One electrode is impure nickel, and the other is the object to be plated. The electrolyzing potential is \(2.50 \mathrm{V}\) (a) Which is the anode and which is the cathode in the experiment? What half- reaction occurs at each electrode? (b) Calculate the number of kilowatt-hours (kWh) of energy required to carry out the electrolysis. \(\left(1 \mathrm{kWh}=3.6 \times 10^{6} \mathrm{J} \text { and } 1 \mathrm{J}=1 \mathrm{C} \times 1 \mathrm{V}\right)\)

Step-by-Step Solution

Answer

(a) Anode is impure nickel; cathode is plated object. (b) Calculate energy using total charge and conversion from Joules to kWh.

Step 1: Determine Electrode Functions

In electrolysis, the anode is where oxidation occurs, and the cathode is where reduction occurs. During nickel plating, nickel atoms from the anode are oxidized and go into solution as Ni²⁺ ions (Ni → Ni²⁺ + 2e⁻). These Ni²⁺ ions are reduced at the cathode and are plated as solid nickel on the object (Ni²⁺ + 2e⁻ → Ni). Therefore, the impure nickel electrode is the anode, and the object to be plated is the cathode.

Step 2: Calculate Surface Area of Cylindrical Object

The surface area of a cylinder is calculated using the formula \(A = 2\pi rh + 2\pi r^2\), where \(r = 2.50 \text{ cm}\) and \(h = 20.00 \text{ cm}\). The additional \(4.0 \text{ mm}\) thickness means \(h\) becomes \(20.40 \text{ cm}\) after plating, but since plating is on the existing surface, we'll find: \[A = 2\pi(2.50)\, \cdot 20.00 + 2\pi(2.50)^2\]\[A \approx 2\cdot 3.14\cdot 2.5\cdot 20 + 2\cdot 3.14\cdot 2.5\cdot 2.5\] which concentrates on the circumference and ends only.

Step 3: Calculate Total Mass of Nickel Required

Using the volume of the coated nickel, \(V = \pi (r + t)^2 h - \pi r^2 h\) where \(r = 2.5 \text{ cm} \), \(t = 0.4 \text{ cm} \) (thickness), and \(h = 20 \text{ cm} \). Calculate the volume and then use the density of nickel (\(\rho_{Ni} = 8.908 \text{ g/cm}^3\)) to find the mass.\[ V = \pi (2.5+0.4)^2 20 - \pi (2.5)^2 20 \] resulting in a mass \(m = V \times \rho_{Ni}\).

Step 4: Calculate Moles of Nickel and Charge Required

From the molar mass of nickel \(58.69 \text{ g/mol}\), find the number of moles of nickel \(\text{mol Ni} = \frac{m}{M(Ni)}\). This is used to find the charge: \( \text{F = 96485} \text{ C/mol} \), with \(2 \text{ electrons per Ni}\) in reduction, thus: \[ \text{Charge} = 2\cdot \text{F}\cdot\text{mol Ni} \].

Step 5: Calculate Energy Required for Electrolysis

With the charge, calculate energy needed using \( E = Q\cdot V \), where \(Q\) is the total charge from Step 4 and \(V = 2.50 \text{ V}\). The energy is first in Joules, then converted to kWh using \(1 \text{ kWh} = 3.6 \times 10^6 \text{ J} \).

Key Concepts

Nickel plating

Nickel plating is a process used to coat a layer of nickel metal onto a surface through electrolysis. This process is often used for corrosion protection, improving aesthetic appeal, or increasing the thickness of the object. In electrolysis, a metal object is submerged in a solution where it acts as one electrode. A nickel bar is often used as another electrode. Through electric current, nickel ions are transferred and coat the object surface. Nickel atoms from the anode lose electrons and form Ni²⁺ ions. These ions migrate through the solution to the cathode, the object being plated, where they gain electrons and deposit as a solid nickel layer. This results in an even and durable nickel coating.

Half-reactions

In electrolysis, half-reactions describe the processes at the anode and cathode. During nickel plating, two important half-reactions occur. The impure nickel acts as the anode and undergoes oxidation. Here, nickel atoms lose electrons:

• Oxidation at the anode: Ni → Ni²⁺ + 2e⁻

The object to be plated serves as the cathode and reduction takes place. Ni²⁺ ions in solution gain electrons and form solid nickel:

• Reduction at the cathode: Ni²⁺ + 2e⁻ → Ni

These half-reactions are fundamental because they summarize the movement of electrons and ions in the plating process. The anode provides the nickel ions, while the cathode is the site of nickel deposition. Through these reactions, the initial material is oxidized, and the ions are transferred to the object for plating.

Cylindrical surface area

The cylindrical surface area plays a vital role in determining the nickel needed for plating. When plating a cylindrical object, calculate its surface area to find out how much nickel is required. The formula for the surface area of a cylinder is:

• Surface Area, A = 2πrh + 2πr²

Here, \(r\) is the radius, and \(h\) is the height of the cylinder. Calculate the lateral surface separately, as plating primarily occurs here. Use the radius and adjusted height, taking into account any plated thickness. In our exercise, before plating, you take measurements of the object, compute the surface area and adjust for the final required nickel coating thickness. This ensures the plated layer covers the object entirely and uniformly.

Energy calculation in electrolysis

The energy calculation in electrolysis is pivotal in determining how much electrical energy is consumed during the process. To compute this, you must first calculate the total charge required. After converting the mass of nickel to moles using its molar mass, determine the charge needed using:

• Total charge, Q = 2 × Faraday's constant (F) × moles of Ni

Where Faraday's constant \(F\) is 96485 C/mol, critical for translating moles to charge. The energy required is given by the equation:

• Energy, E = Q × voltage (V).

In this case, use V = 2.50 V for the electrolyzing potential. The energy initially calculated in Joules can be converted to kilowatt-hours (kWh) for practical understanding using the conversion factor \(1 ext{ kWh} = 3.6 imes 10^6 ext{ J}\). Hence, understanding this aspect ensures efficient use of electricity, aiding in correct energy management throughout the electrolysis process.