Key Concept: Electroplating, Current Density
a) 0.1325 cm
[Solution Description]
First, calculate the total charge passed during the process. Charge (Q) is given by $Q = I \times t$, where $I$ is the current and $t$ is the time. Current density ($J$) is given by $J = I/A$, where $A$ is the area. Rearranging, $I = J \times A$. Substituting the values, $I = 0.5 \, \text{A/cm}^2 \times 10 \, \text{cm}^2 = 5 \, \text{A}$. Time ($t$) is 2 hours = 7200 seconds. So, $Q = 5 \, \text{A} \times 7200 \, \text{s} = 36000 \, \text{C}$.
Next, use Faraday's law to find the mass of gold deposited. Faraday's law is given by $m = \frac{Q \times M}{n \times F}$, where $M$ is the molar mass, $n$ is the valency, and $F$ is Faraday's constant (96500 C/mol). Substituting the values, $m = \frac{36000 \, \text{C} \times 197 \, \text{g/mol}}{3 \times 96500 \, \text{C/mol}} \approx 25.6 \, \text{g}$.
Finally, calculate the thickness of the gold layer. Volume ($V$) is given by $V = m/\rho$, where $\rho$ is the density. Substituting the values, $V = 25.6 \, \text{g} / 19.32 \, \text{g/cm}^3 \approx 1.325 \, \text{cm}^3$. Since volume is also given by $V = A \times d$, where $d$ is the thickness, $d = V/A = 1.325 \, \text{cm}^3 / 10 \, \text{cm}^2 \approx 0.1325 \, \text{cm}$.
Your Answer is correct.
a) 0.1325 cm
[Solution Description]
First, calculate the total charge passed during the process. Charge (Q) is given by $Q = I \times t$, where $I$ is the current and $t$ is the time. Current density ($J$) is given by $J = I/A$, where $A$ is the area. Rearranging, $I = J \times A$. Substituting the values, $I = 0.5 \, \text{A/cm}^2 \times 10 \, \text{cm}^2 = 5 \, \text{A}$. Time ($t$) is 2 hours = 7200 seconds. So, $Q = 5 \, \text{A} \times 7200 \, \text{s} = 36000 \, \text{C}$.
Next, use Faraday's law to find the mass of gold deposited. Faraday's law is given by $m = \frac{Q \times M}{n \times F}$, where $M$ is the molar mass, $n$ is the valency, and $F$ is Faraday's constant (96500 C/mol). Substituting the values, $m = \frac{36000 \, \text{C} \times 197 \, \text{g/mol}}{3 \times 96500 \, \text{C/mol}} \approx 25.6 \, \text{g}$.
Finally, calculate the thickness of the gold layer. Volume ($V$) is given by $V = m/\rho$, where $\rho$ is the density. Substituting the values, $V = 25.6 \, \text{g} / 19.32 \, \text{g/cm}^3 \approx 1.325 \, \text{cm}^3$. Since volume is also given by $V = A \times d$, where $d$ is the thickness, $d = V/A = 1.325 \, \text{cm}^3 / 10 \, \text{cm}^2 \approx 0.1325 \, \text{cm}$.
