Question

how many minutes will it take to plate the spoon with 0.25 g silver? $\ce{ag+ + e- -> ag}$ $e_{\text{red}}^{\circ} = 0.80$ v $i = 4.0$ amp $t = ?$ min

Explanation:

Step1: Calculate moles of Ag

Molar mass of Ag is 107.87 g/mol. Moles of Ag, \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.25\ \text{g}}{107.87\ \text{g/mol}} \approx 0.002318\ \text{mol} \)

Step2: Determine moles of electrons

From the reaction \( \text{Ag}^+ + \text{e}^- ightarrow \text{Ag} \), 1 mole of Ag is produced per mole of electrons. So moles of electrons, \( n_{e^-} = 0.002318\ \text{mol} \)

Step3: Calculate charge (Q)

Using Faraday's constant \( F = 96485\ \text{C/mol} \), \( Q = n_{e^-} \times F = 0.002318\ \text{mol} \times 96485\ \text{C/mol} \approx 223.6\ \text{C} \)

Step4: Relate charge to current and time

Current \( I = \frac{Q}{t} \), so \( t = \frac{Q}{I} \). \( I = 4.0\ \text{A} \) (since 1 A = 1 C/s), so \( t = \frac{223.6\ \text{C}}{4.0\ \text{C/s}} = 55.9\ \text{s} \)

Step5: Convert seconds to minutes

\( t = \frac{55.9\ \text{s}}{60\ \text{s/min}} \approx 0.93\ \text{min} \)? Wait, no, wait, I think I made a mistake in moles. Wait, let's recalculate moles: \( 0.25 / 107.87 ≈ 0.002318 \) moles. Then charge is \( 0.002318 \times 96485 ≈ 223.6 \) C? Wait, no, 0.002318 * 96485 is approximately 223.6? Wait, 0.002 * 96485 = 192.97, 0.000318*96485≈30.6, total≈223.57. Then time in seconds: 223.57 / 4 ≈ 55.89 seconds. Then minutes: 55.89 / 60 ≈ 0.93 minutes? That can't be right. Wait, no, wait, maybe I messed up the reaction. Wait, the spoon is the cathode, so Ag+ is reduced to Ag. The anode is Ag, which oxidizes to Ag+. Wait, but the amount of Ag plated is 0.25 g. Wait, maybe my calculation of moles is wrong. Wait, 0.25 grams of Ag: molar mass 107.87, so 0.25 / 107.87 ≈ 0.002318 moles. Then electrons: 1 mole e- per mole Ag, so moles of e- is 0.002318. Charge: 0.002318 * 96485 = 223.6 C. Then time: Q = I*t, so t = Q/I = 223.6 / 4 = 55.9 seconds, which is ~0.93 minutes. But that seems too short. Wait, maybe I made a mistake in the mass. Wait, 0.25 grams is 250 milligrams. Wait, maybe the current is 4.0 amps, which is 4 C/s. Wait, let's check again. Wait, no, wait, 0.25 g Ag: moles = 0.25 / 107.87 ≈ 0.002318 mol. Charge = 0.002318 mol * 96485 C/mol = 223.6 C. Time = 223.6 C / 4 C/s = 55.9 s = 55.9 / 60 ≈ 0.93 min. But that seems too fast. Wait, maybe the current is 4.0 amps, which is 4 coulombs per second. Wait, maybe I messed up the Faraday's law. Wait, Faraday's law: \( m = \frac{M \times I \times t}{n \times F} \), where m is mass, M is molar mass, I is current, t is time, n is moles of electrons per mole of product. Here, n=1 (since Ag+ + e- -> Ag, so n=1). So rearranging for t: \( t = \frac{m \times n \times F}{M \times I} \). Let's use this formula. So m=0.25 g, n=1, F=96485 C/mol, M=107.87 g/mol, I=4.0 A. So t = (0.25 g * 1 * 96485 C/mol) / (107.87 g/mol * 4.0 C/s) = (24121.25) / (431.48) ≈ 55.9 seconds. Then convert to minutes: 55.9 / 60 ≈ 0.93 minutes. Wait, that seems correct? Wait, 4 amps is 4 coulombs per second. So in 1 minute, 4*60=240 coulombs. So 223 coulombs is less than 240, so about a minute. Wait, maybe the problem is correct. Alternatively, maybe I made a mistake in molar mass. Wait, Ag molar mass is 107.8682 g/mol. So 0.25 / 107.8682 ≈ 0.002318 mol. Then charge is 0.002318 * 96485 ≈ 223.6 C. Time is 223.6 / 4 = 55.9 s ≈ 0.93 min. But that seems too short. Wait, maybe the current is 4.0 milliamps? No, the problem says 4.0 amp. Wait, maybe I messed up the reaction. Wait, the anode is Ag, so Ag -> Ag+ + e-, and cathode is Ag+ + e- -> Ag. So it's an electrolytic cell for plating? Wait, no, if the anode is Ag and cathode is spoon, then Ag from anode dissolves, and Ag+ from solution plates onto cathode. So the amount of Ag plated is equal to the amount of Ag dissolved. But the problem is asking to plate 0.25 g Ag. So the calculation should be correct. Wait, maybe the answer is approximately 0.93 minutes, or maybe I made a mistake in the number of moles. Wait, let's recalculate: 0.25 grams of Ag. Molar mass 107.87 g/mol. So moles = 0.25 / 107.87 = 0.002318 mol. Then, since each Ag+ gains 1 electron, moles of electrons = 0.002318 mol. Charge Q = n * F = 0.002318 * 96485 = 223.6 C. Current I = 4.0 A = 4.0 C/s. Time t = Q / I = 223.6 / 4 = 55.9 s. Convert to minutes: 55.9 / 60 = 0.93 min. So the answer is approximately 0.93 minutes. Wait, but maybe the problem expects a different approach. Wait, maybe I made a mistake in Faraday's constant. Wait, Faraday's constant is 96485 C/mol. So that's correct. Alternatively, maybe the mass is 0.25 kg? No, the problem says 0.25 g. So I think the calculation is correct. So the time is approximately 0.93 minutes. But let's check with another method. Let's use the formula for electroplating: \( t = \frac{m \times n \times F}{M \times I} \). Plugging in: m=0.25 g, n=1, F=96485, M=107.87, I=4.0. So t = (0.25 * 1 * 96485) / (107.87 * 4.0) = (24121.25) / (431.48) ≈ 55.9 seconds, which is 55.9/60 ≈ 0.93 minutes. So the answer is approximately 0.93 minutes. Wait, but maybe the problem has a typo, or I misread the current. Wait, 4.0 amp is 4000 milliamps, which is a lot. So plating 0.25 g in less than a minute with 4 amps is possible? Let's see, 4 amps is 4 coulombs per second. In one minute, 240 coulombs. 240 coulombs is 240 / 96485 ≈ 0.002487 moles of electrons, which is 0.002487 moles of Ag, which is 0.002487 * 107.87 ≈ 0.268 g. So that's close to 0.25 g. So the time should be a bit less than a minute, which matches our calculation. So the time is approximately 0.93 minutes, or about 56 seconds. So the answer is approximately 0.93 minutes.

Answer:

\boxed{0.93} (minutes, approximately)