Which of the following ions  requires the largest quantity of electricity for discharge at an electrode?

Analysis

To determine which ion requires the largest quantity of electricity for discharge, we use Faraday’s First Law of Electrolysis. The quantity of electricity ($Q$) required to discharge an ion is directly proportional to the number of moles of the ion and its valency (charge).

Concept

The formula for the quantity of electricity ($Q$) in terms of Faraday's constant ($F$) is:

Where:

• $n$ = number of moles of the ion
• $z$ = charge (valency) of the ion
• $F$ = Faraday's constant (approx. 96,500 Coulombs/mole)

Since $F$ is a constant, we only need to calculate the product of moles ($n$) × charge ($z$) for each option to compare the relative quantities of electricity.

Step-by-Step Solution

Let's calculate the value of $n \times z$ for each option:

• Option A: 2.0 mole of Q³⁺ • $n = 2.0$, $z = 3$ • $Q \propto 2.0 \times 3 = \mathbf{6.0}$

• Option B: 2.5 mole of R²⁺ • $n = 2.5$, $z = 2$ • $Q \propto 2.5 \times 2 = \mathbf{5.0}$

• Option C: 3.0 mole of T⁻ • $n = 3.0$, $z = 1$ • $Q \propto 3.0 \times 1 = \mathbf{3.0}$

• Option D: 4.0 mole of Y⁻ • $n = 4.0$, $z = 1$ • $Q \propto 4.0 \times 1 = \mathbf{4.0}$

Comparing the results (6.0, 5.0, 3.0, and 4.0), the highest value is 6.0, which corresponds to Option A.

Conclusion

Option A requires the largest quantity of electricity because it involves the highest product of moles and ionic charge.

Exam Tip

When solving electrolysis problems, remember that the charge of the ion acts as a "multiplier" for the amount of electricity needed. Even if you have a large number of moles, a low-valency ion (like a monovalent ion with charge 1) might still require less electricity than a smaller number of moles of a highly charged ion (like a trivalent ion with charge 3). Always calculate $n \times z$ to be safe!