In an electrolysis experiment a student passes the same quantity of electricity through

Question

From the following information, calculate the solubility product of AgBr:

Answer 1

Equivalent weight of Ag = \(\frac{108}1 = 108 g/equivalent\)

Number of faradays = \(\frac{6.97}{108} = 0.0645 F\)

Let say, oxidation state of Au = +n

∴ Equivalent weight of Au = \(\frac {197}{n} g/equivalent\)

∴ Number of faraday required to deposite 4.25g of Au = \(\frac{4.25\times n}{197}\)

\(\therefore \frac{4.25\times n}{197} = 0.0645\)

⇒ \(n = \frac{0.0345\times 197}{4.25}\)

\(n = 2.989\)

\(n \approx +3\)

Hence oxidation state of Au is +3.