You conduct another experiment on silver condensation. You add 8.0 g of silver nitrate to 200 mL of pure water. You attach a single 850 mAh AAA battery (at 1.3 V) to the electrochemical cell. You can assume that the entire battery charge is useable for electroplating. You get 3.24 g of silver metal at the end of the experiment.
a) Was your silver recovery limited by the battery capacity or the available silver in solution?
b) What was your product yield (%) using the limiting resource?
See above.
To determine whether the silver recovery was limited by the battery capacity or the available silver in solution, we need to compare the amount of silver obtained in the experiment with the maximum amount that could have been obtained using each resource.
a) To determine whether the battery capacity was the limiting factor, we need to calculate the maximum amount of silver that could have been obtained using the available charge in the battery. We know that the battery's capacity is 850 mAh (milliampere-hour) and its voltage is 1.3 V. By multiplying the capacity by the voltage, we can calculate the maximum charge available in the battery in units of mAh⋅V. In this case, it would be:
Maximum charge available in the battery = 850 mAh * 1.3 V = 1105 mAh⋅V
Next, we need to convert this charge into coulombs (C). Since 1 coulomb (C) is equivalent to 3600 milliampere-seconds (mA⋅s), we can calculate the charge in coulombs by dividing the available charge in mAh⋅V by 3600:
Maximum charge available in the battery (C) = 1105 mAh⋅V / 3600 = 0.307 C
Since 1 Faraday (F) is equivalent to 96485 C and we know that 1 mole of electrons corresponds to 1 Faraday, we can use this conversion to determine the maximum amount of silver that could have been deposited by the available charge in the battery.
The molar mass of silver (Ag) is 107.87 g/mol, so we can calculate the maximum amount of silver that could have been obtained using the available charge:
Maximum amount of silver (g) = (Maximum charge available in the battery (C) * molar mass of silver (g/mol)) / (Faraday's constant (C/mol))
Plugging in the values, we get:
Maximum amount of silver (g) = (0.307 C * 107.87 g/mol) / (96485 C/mol) = 0.000342 g
Comparing the maximum amount of silver (0.000342 g) that could have been obtained using the battery with the actual amount obtained (3.24 g), we can conclude that the silver recovery was not limited by the battery capacity.
Therefore, the silver recovery was limited by the available silver in solution.
b) To calculate the product yield using the limiting resource, we divide the actual amount of silver obtained (3.24 g) by the maximum amount of silver that could have been obtained using the limiting resource (which we determined to be 0.000342 g):
Product yield (%) = (Actual amount of silver obtained (g) / Maximum amount of silver obtained using limiting resource (g)) * 100
Plugging in the values, we get:
Product yield (%) = (3.24 g / 0.000342 g) * 100 ≈ 947.37%
The product yield is approximately 947.37%.