95. In an electroplating setup, if 2.5 g of silver is deposited on an ornament in 1 hour, what current was used? (Atomic mass of silver = 108 g/mol, valency = 1, Faraday's constant = 96500 C/mol)
Key Concept: Electroplating, Faraday's Law
a) 0.62 A
[Solution Description]
Use Faraday's law to find the charge required to deposit 2.5 g of silver. Faraday's law is given by $m = \frac{Q \times M}{n \times F}$, rearranging for charge ($Q$), $Q = \frac{m \times n \times F}{M}$. Substituting the values, $Q = \frac{2.5 \, \text{g} \times 1 \times 96500 \, \text{C/mol}}{108 \, \text{g/mol}} \approx 2234.26 \, \text{C}$.
Next, calculate the current ($I$) using $Q = I \times t$, where $t$ is the time in seconds. Time ($t$) is 1 hour = 3600 seconds. So, $I = Q / t = 2234.26 \, \text{C} / 3600 \, \text{s} \approx 0.62 \, \text{A}$.
Your Answer is correct.
a) 0.62 A
[Solution Description]
Use Faraday's law to find the charge required to deposit 2.5 g of silver. Faraday's law is given by $m = \frac{Q \times M}{n \times F}$, rearranging for charge ($Q$), $Q = \frac{m \times n \times F}{M}$. Substituting the values, $Q = \frac{2.5 \, \text{g} \times 1 \times 96500 \, \text{C/mol}}{108 \, \text{g/mol}} \approx 2234.26 \, \text{C}$.
Next, calculate the current ($I$) using $Q = I \times t$, where $t$ is the time in seconds. Time ($t$) is 1 hour = 3600 seconds. So, $I = Q / t = 2234.26 \, \text{C} / 3600 \, \text{s} \approx 0.62 \, \text{A}$.
