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During the electrolysis of a salt of a metal M, a current of 0.5 A flows for 32 minutes and 10 seconds, depositing 0.325 g of M. What is the charge of the metal ion?

(M = 65, \(F\) = 96,500 C per mole of electrons)

A. 1
B. 2
C. 3
D. 4

Answer :

The charge of the metal ion M is determined to be 2 based on the calculations involving current, time, and Faraday's constant, leading to the conclusion that the answer is B. 2.

Step-by-Step Solution

1. First, calculate the total charge (Q) passed during the electrolysis. The current (I) is 0.5 A and the time (t) is 32 minutes and 10 seconds.

  • Convert the time to seconds: t = 32 * 60 + 10 = 1930 seconds
  • Q = I * t = 0.5 A * 1930 s = 965 Coulombs

2. Next, determine the moles of metal M deposited. Given that the molar mass of M is 65 g/mol:

  • Moles of M (n) = mass / molar mass = 0.325 g / 65 g/mol = 0.005 mol

3. Calculate the charge per mole of electrons (Faraday's constant): 96,500 C/mol

4. Find the number of moles of electrons required to deposit 0.005 mol of M:

965 Coulombs = n * Faraday's constant * moles of electrons

  • n * 96,500 C/mol * x mol electrons = 965 C
  • n * x = 0.01

5. The answer is the charge of the metal ion, which corresponds to the number of moles of electrons per mole of metal deposited:

For n = 1, the charge is 1. For n = 2, the charge is 2, and so on.

Thus, the charge of the metal ion M is 2.

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