$\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)}$
Correct Answer: 2
Approach Solution - 1
1. **Understanding the Reaction**:
The electrochemical reaction at the cathode involves the reduction of AuCl4- ions to gold (Au), where each ion gains 3 electrons:
$$\text{AuCl}_4^- + 3e^- \rightarrow \text{Au} + 4\text{Cl}^-$$
2. **Calculate Moles of Au Deposited**:
The molar mass of Au is given as 197 g/mol. The mass of Au deposited is 1.314 g. The number of moles of Au deposited is:
$$\text{Moles of Au} = \frac{1.314 \text{ g}}{197 \text{ g/mol}} = 0.00667 \text{ mol}$$
3. **Electrons Transferred**:
Since each Au atom corresponds to the transfer of 3 electrons, the total moles of electrons (n) is:
$$n = \text{Moles of Au} \times 3 = 0.00667 \times 3 = 0.02001 \text{ mol e}^-$$
4. **Calculate Total Charge**:
The total charge (Q) transferred is given by Faraday's constant (F = 96485 C/mol), where:
$$Q = n \times F = 0.02001 \times 96485 = 1930.34985 \text{ C}$$
5. **Express Charge in Faradays**:
Since 1 Faraday (F) is 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. **Express in Given Format**:
The charge expressed in the given format is:
$$0.02002 \times 10^{-2} \text{ F} = 2.002 \times 10^{-2} \text{ F}$$
7. **Verification Against Range**:
The computed charge, 2.002, indeed falls within the given range of 2 to 2.
Thus, the total charge passed through the solution is 2.002 × 10-2 F.
Approach Solution -2
Solution:
\(\frac{W}{E} = \frac{\text{charge}}{1F}\)
Substitute the values:
\(\frac{1.314}{\frac{197}{3}} = \frac{Q}{1F}\)
\(Q = 2 \times 10^{-2} F\)
The Correct Answer is: 2
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