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

Next, we need to calculate the total charge transferred using Faraday's law of electrolysis. One faraday corresponds to the charge needed to transfer one mole of electrons (approximately 96485 C). The total charge transferred when 0.019 faradays are passed through is:

$$ext{Total Charge} = 0.019 \, ext{faradays} \times 96485 \, ext{C/faraday} \approx 1831.36 \, ext{C}$$

To find the number of moles of electrons used in this process, we can divide the total charge by the Faraday constant:

$$ext{Moles of Electrons} = \frac{1831.36 \, ext{C}}{96485 \, ext{C/mol}} \approx 0.0190 \, ext{mol}$$

Since each chromium atom corresponds to either +2, +3, +4, or +6 oxidation states, we can establish a relationship between the moles of chromium produced and the moles of electrons:

If each chromium atom takes x electrons, then:

$$\text{Moles of Cr} = \frac{\text{Moles of Electrons}}{x}$$

For the total of 0.00962 moles of chromium:

• If x = 3 (Cr in +3 state),

$$0.00962 = \frac{0.0190}{3} \Rightarrow 0.00962 \approx 0.00633 \, [ ext{not compatible}]$$

• If x = 2 (Cr in +2 state),

$$0.00962 = \frac{0.0190}{2} \Rightarrow 0.00962 \approx 0.0095 \, [ ext{compatible}]$$

Hence, the oxidation number of chromium in the compound is most likely +2.