An ornament was coated with a metal, M, by electrolysis of a solution of the metal ion, M²⁺ - VCE - SSCE Chemistry - Question 19 - 2011 - Paper 1

Using Faraday's law, we can calculate the moles of electrons:

$n_e = \frac{Q}{F}$

where:

• F (Faraday's constant) = 96500 C/mol

Substituting the values gives:

$n_e = \frac{270 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.00280 \, \text{mol}$

From the electrolysis, we know that:

$n = \frac{n_{metal}}{x}$

where:

• n_metal = 0.0014 mol (given)
• x = number of electrons transferred per ion reaction (to be determined)
• n_e = moles of electrons calculated above

Rearranging gives:

$x = \frac{n_{metal}}{n_e} = \frac{0.0014}{0.00280} = 0.5$

Since M²⁺ implies that each ion requires 2 electrons to be reduced to metal, we have:

$x = 2$

Thus, the value of x in M^x+ is 2, making the answer B.