The passage of 0.019 faradays of electricity through a molten chromium compound yields 0.50 g of chromium metal - VCE - SSCE Chemistry - Question 8 - 2006 - Paper 1

Using Faraday's law of electrolysis, the number of moles of electrons (n) transferred can be defined as:

$n = \text{Faradays} \cdot 1 \text{ mol/electrons} = 0.019 \text{ faradays}$

Since 1 faraday corresponds to 1 mole of electrons, the total charge is 0.019 moles of electrons.

In electrolysis, the relationship between the moles of chromium reduced and the moles of electrons can be expressed by the oxidation state of chromium (n). The relation is given by:

$\text{Moles of electrons} = n \times \text{moles of Cr}$

Substituting the known values:

$0.019 = n \times 0.00962$

From which we derive:

$n \approx \frac{0.019}{0.00962} \approx 1.975 \approx 2$

The oxidation number of chromium is thus likely to be +2 when it is reduced from its compound state to the metallic form. This is supported by the stoichiometric relationship defined before.

Considering the calculations and relationships, the oxidation number of chromium in the compound is likely to be +2.