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 for the deposition of Ni is 60%.

Step 1: Calculate the Moles of Ni Deposited

To calculate the moles of Ni deposited, use the formula \( n = \frac{m}{M} \), where \( m \) is the mass and \( M \) is the molar mass. Given \( m = 9.9\, \text{g} \) and \( M = 58.7 \), we have \( n = \frac{9.9}{58.7} \approx 0.1686\, \text{moles} \).

Step 2: Calculate the Charge for Ni Deposition

The reaction for nickel deposition is \( \text{Ni}^{2+} + 2\,e^- \rightarrow \text{Ni} \). Therefore, 2 moles of electrons are needed per mole of Ni. The total charge required is \( 2 \times 0.1686 \times 96485 \approx 32536\, \text{Coulombs} \).

Step 3: Total Charge Supplied to the System

The total charge is given by \( Q = I \cdot t \), where current \( I = 15\, \text{A} \) and time \( t \) in seconds is needed. First, calculate moles of \( \text{H}_2 \) produced using \( \text{moles} = \frac{2.51}{22.4} \approx 0.112\, \text{moles} \). Since \( 2 \, \text{H}^{+} + 2\, e^- \rightarrow \text{H}_2 \), it requires \( 0.112 \times 2 \times 96485 \approx 21622\, \text{Coulombs} \) for \( \text{H}_2 \) production. Thus, total charge \( Q_{total} = 32536 + 21622 = 54158\, \text{Coulombs} \).

Step 4: Calculate Current Efficiency

Current efficiency \( \eta \) is calculated as \( \eta = \left( \frac{\text{Charge for Ni}}{\text{Total Charge}} \right) \times 100 \). Thus, \( \eta = \left( \frac{32536}{54158} \right) \times 100 \approx 60\% \).

Step 5: Conclusion

Based on the calculation of current efficiency, the answer is option (a) \( 60\% \).

Key Concepts

Nickel Plating

Nickel plating is an electrochemical process used widely for depositing a layer of nickel onto a metal surface. This enhances the appearance and provides corrosion resistance. Nickel plating generally takes place in a \( \mathrm{NiSO}_4 \) bath, where nickel ions are reduced at the cathode.

When an electric current flows through the bath, nickel ions in the solution gain electrons and deposit as a solid metal layer on the substrate or part being plated. The improvement in protective and aesthetic qualities makes it a popular choice in industries such as automotive and electronics.

• The cathode is where the plating happens, in this example, nickel and hydrogen are deposited/released.
• Heating and agitation of the bath often optimize plating conditions.
• Impurities in the bath can impact the quality and consistency of the nickel plating.

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis explain the relationship between the electric current used in the electrolysis process and the amount of substance that gets deposited at the electrode.

The First Law states that the amount of substance deposited is directly proportional to the amount of electricity (charge) passed through the solution. This means that as more electric current flows, more material gets deposited. The Second Law states that the mass of substances deposited by passing the same quantity of electricity is proportional to their equivalent weights.

• This is crucial in understanding current efficiency, as it allows us to calculate how much of the charge contributes to the actual plating process.
• Faraday's constant,\( 96485 \ \text{C/mol} \), is used to quantify this relationship.
• It is foundational in calculating how much electrical charge is needed to deposit a specific mass of nickel.

Moles Calculation

Knowing how to calculate moles is pivotal in determining how much nickel is deposited during electrolysis. Using the formula \( n = \frac{m}{M} \), we calculate the moles \( n \), where \( m \) is the mass of nickel deposited, and \( M \) is the molar mass.

For nickel, the molar mass is given as \( 58.7 \ \text{g/mol} \), and if \( 9.9 \ \text{g} \) of nickel is deposited, then:\[ n = \frac{9.9}{58.7} \approx 0.1686 \ \text{moles} \]

• The calculation of moles is the first part of finding out how much charge is needed.
• Moles calculation precedes the charge calculation step, it gives a clear count of the nickel atoms deposited.
• This step is vital for both practical and theoretical purposes, ensuring accurate results in any electrochemical application.

Coulombs Calculation

Coulombs calculation is about quantifying the total electric charge used during the electrolysis process.

The formula \( Q = I \cdot t \), where \( I \) is the current in amperes and \( t \) is time in seconds, helps determine the total charge. However, we also use the conversion in terms of electrons required to deposit a specific amount of nickel or produce hydrogen gas.

- For nickel deposition: Two moles of electrons are required per mole of nickel.- For hydrogen evolution: Two moles of electrons reduce one mole of hydrogen gas.Understanding the total charge involved helps in computing current efficiency, which eventually tells us how efficiently the current is being used for nickel deposition rather than other side reactions, like hydrogen gas production.

• Knowing the total charge used for nickel and hydrogen helps isolate efficiencies and losses, clarifying the effectiveness of the plating process.
• Faraday's constant \( 96485 \ \text{C/mol} \) is instrumental here, relating electron involvement to macroscopic charge calculations.
• Determining total Coulombs used is key for optimizing processes in industrial applications.