Question

Consider the following half-cell reaction:

\[\ce{Au^{3+} + 3e^- -> Au_{(s)}}\]

How many coulombs of electricity is required to deposits 0.394 g of Au.

(Given - Molar mass of Au = 197 g mol⁻¹)

विकल्प

• 289.5 C

• 579.0 C

• 386.0 C

• 193.0 C

Answer 1

579.0 C

Explanation:

\[\ce{Au^{3+} + 3e^- -> Au_{(s)}}\]

Charge required to produce 197 gm of Au = 3F.

Charge required to produce 0.394 gm of Au = ?

= `(0.394 xx 3 "F")/197` = 0.006 F = 0.006 × 96500 C = 579 C