$\text{A constant current was passed through a solution of AuCl}_4^- \text{ ion between gold electrodes. After a} \\\text{period of 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed} \\$$\text{through the solution is \_\_\_\_\_\_\_\_} \times 10^{-2} \, \text{F.} \\$$\text{(Given atomic mass of Au = 197)}$
To ascertain the total charge transferred through the solution, Faraday's laws of electrolysis are utilized. The procedure is as follows:
1. Reaction Analysis: At the cathode, AuCl4- ions are reduced to gold (Au), with each ion accepting 3 electrons: $$\text{AuCl}_4^- + 3e^- \rightarrow \text{Au} + 4\text{Cl}^-$$
2. Gold Moles Calculation: Given the molar mass of Au as 197 g/mol and the deposited mass as 1.314 g, the moles of Au deposited are: $$\text{Moles of Au} = \frac{1.314 \text{ g}}{197 \text{ g/mol}} = 0.00667 \text{ mol}$$
3. Electron Transfer Moles: Each gold atom deposited signifies the transfer of 3 electrons. Therefore, the total moles of electrons (n) transferred is: $$n = \text{Moles of Au} \times 3 = 0.00667 \times 3 = 0.02001 \text{ mol e}^-$$
4. Total Charge Calculation: Using Faraday's constant (F = 96485 C/mol), the total charge (Q) transferred is: $$Q = n \times F = 0.02001 \times 96485 = 1930.34985 \text{ C}$$
5. Charge in Faradays: As 1 Faraday (F) equals 96485 Coulombs, the charge in Faradays is: $$\text{Charge in F} = \frac{1930.34985 \text{ C}}{96485 \text{ C/F}} = 0.02002 \text{ F}$$
6. Format Adjustment: The charge, adjusted to the specified format, is: $$0.02002 \times 10^{-2} \text{ F} = 2.002 \times 10^{-2} \text{ F}$$
7. Range Verification: The calculated charge value, 2.002, is confirmed to be within the specified range of 2 to 2.
Consequently, the total charge transmitted through the solution is 2.002 × 10-2 F.
