Problem 144 - A Complete Resource Book in Chemistry for JEE Main

Question

A current of 15 amp is employed to plate Nickel in a \(\mathrm{NiSO}_{4}\) bath. Both \(\mathrm{Ni}\) and \(\mathrm{H}_{2}\) are formed at the cathode. If \(9.9 \mathrm{~g}\) of \(\mathrm{Ni}\) are deposited with the simultaneous liberation of \(2.51\) litres of \(\mathrm{H}_{2}\) measured at STP, what is the current efficiency for the deposition of Ni? (Atomic weight of \(\mathrm{Ni}=58.7\) ) (a) \(60 \%\) (b) \(70 \%\) (c) \(80 \%\) (d) \(56 \%\)

Step-by-Step Solution

Answer

The current efficiency is approximately 60.08%, so the answer is (a) 60%.

Step 1: Calculate Moles of Nickel Deposited

First, calculate the number of moles of Nickel (\( \text{Ni} \)) deposited using the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Given \( \text{mass} = 9.9 \text{ g} \) and \( \text{molar mass of Ni} = 58.7 \text{ g/mol} \), the calculation is: \[ \text{moles of Ni} = \frac{9.9}{58.7} \approx 0.1686 \text{ moles} \]

Step 2: Determine Charge Required for Nickel Deposition

The reaction at the cathode for nickel deposition is \( \text{Ni}^{2+} + 2\text{e}^- \rightarrow \text{Ni} \). This means 2 moles of electrons are needed to deposit 1 mole of \( \text{Ni} \). Calculate the total charge required for the deposited \( \text{Ni} \):\[ \text{Charge (C)} = \text{moles of Ni} \times 2 \text{ moles of e}^- \times 96500 \text{ C/mol e}^- \]Substituting the values, we get:\[ \text{Charge (C)} = 0.1686 \times 2 \times 96500 \approx 32535 \text{ C} \]

Step 3: Calculate Total Charge Passed

The total charge passed through the system is calculated by multiplying the current and the time in seconds. Given the current \( I = 15 \text{ A} \) and time \( t \) such that \( 2.51 \text{ L of H}_2 \) is liberated:To find \( t \), we use the reaction: \( 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \). Thus, \[ \frac{2}{22.4} \text{ moles of } \text{H}_2 \Rightarrow 2 \times \frac{2.51}{22.4} \times 96500 \approx 21632.59 \text{ C} \]Total charge passed \( Q = I \times t = 54167.59 \text{ C} \) based on substitution \( 7616.35\text{ C}+21632.59 \text{ C} \).

Step 4: Calculate Current Efficiency

Current efficiency is the ratio of the charge used for nickel deposition to the total charge passed, expressed as a percentage:\[ \text{Current Efficiency} = \left( \frac{32535}{54167.59} \right) \times 100 \% \approx 60.08 \% \]

Step 5: Find the Correct Option

Based on the calculated current efficiency of approximately \( 60.08\% \), the closest option is (a) \(60 \%\).

Key Concepts

Current efficiency

When we talk about electroplating, current efficiency is a crucial factor. It tells us how effectively the electric charge contributes to the intended deposition of metal on a surface. In electroplating, especially with nickel, current efficiency is essentially the ratio of the charge used for depositing the metal to the total charge that passes through the circuit. It is expressed as a percentage:

• When electric current flows, not only desired metal cations get deposited but other reactions might also occur, such as hydrogen evolution.
• The goal is to ensure a high percentage of the total current is used for the nickel deposition and not wasted on side reactions.

For instance, in a scenario where we are depositing nickel and we're given a current efficiency of about 60%, this means that 60% of the total electric charge was effectively used to deposit nickel, while the rest was part of other reactions, like hydrogen production. The efficiency can be calculated using the formula:\[\text{Current Efficiency} = \left( \frac{\text{Charge used for Nickel}}{\text{Total Charge Passed}} \right) \times 100\%\]

Nickel deposition

Nickel deposition is a particular focus in electroplating. It involves the reduction of nickel ions in an electrolytic solution to form a solid nickel layer on a substrate. In our case with a \( \text{NiSO}_4 \) solution:

• The process occurs at the cathode where the positively charged \( \text{Ni}^{2+} \) ions are attracted.
• Via electrochemical reactions, these ions gain two electrons each to become neutral nickel atoms which then form a thin layer.

This conversion from \( \text{Ni}^{2+} \) ions to nickel metal is crucial. Every two moles of electrons are needed to deposit one mole of nickel. Therefore, if more nickel is needed, more charge is necessary, following Faraday's law of electrolysis.To ensure uniform nickel thickness:

• The current density must be managed.
• The solution should be evenly mixed.
• Temperature should be kept constant during the process.

For successful nickel plating, controlling these variables is essential, especially in industrial applications where surface quality and durability matter.

Electrochemistry

Electrochemistry is the branch of chemistry that explores the relationship between electricity and chemical change. When we delve into electroplating:

• It involves understanding how electricity can facilitate chemical transformations.
• Specifically, how electric current drives the reduction of metal ions at the cathode.

In the context of our nickel plating exercise, electrochemistry explains:

• The movement of \( \text{Ni}^{2+} \) ions towards the cathode.
• Their acceptance of electrons to become solid nickel metal.

The basic principles include reactions at electrodes:

• At the cathode (negative electrode), reduction reactions occur where electrons are gained, depositing nickel.
• At the anode, often oxidation reactions occur where electrons are lost.

Electrochemistry allows us to calculate:

• Current efficiencies.
• Predict how much metal will deposit for a given charge.
• Monitor side reactions such as hydrogen formation.

Understanding electrochemistry empowers engineers and chemists to optimize electroplating processes for efficiency and quality.

Faraday's laws of electrolysis

Faraday's laws of electrolysis are fundamental principles that define the quantitative aspects of electrochemical reactions. These laws state:

• The amount of substance deposited or dissolved at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
• The amount of chemical change is also proportional to the equivalent weight of the substance.

In nickel deposition:

• The first law allows us to calculate the charge required to deposit a specific mass of nickel.
• The second law can compare the deposition of different substances under similar electrical conditions.

Mathematically, Faraday's first law is expressed as:\[\text{Mass deposited} = \frac{Q \cdot E}{F}\]Where:

• \( Q \) is the total electric charge.
• \( E \) is the equivalent weight of nickel.
• \( F \) is Faraday's constant, approximately 96500 C/mol.

These laws are critical in designing processes to achieve the desired thickness and quality of electro-deposited layers. They also help in calculating the current efficiency by comparing the actual deposited mass with the theoretical prediction based on the input electricity.