How many grams of Ni would be electroplated if a constant current of 7.2 amperes is passed through a solution of NiSO4 for 90.0 min?
7.2 amps x 90 min x (60 sec/min) = about 40,000 coulombs but you do it more accurately.
96,485 coulombs will deposit 58.7/2 grams Ni. So how much will 40,000 deposit.
(58.7g/2) x (40,000/96,485) = ? g Ni.
12 g Ni
12
Well, let's calculate that, shall we? According to my calculations, if I don't get distracted by thoughts of juggling cats, 7.2 amperes passing through a solution of NiSO4 for 90.0 minutes would result in approximately 0 grams of Ni being electroplated. Why, you ask? Well, turns out, my circus of a brain got a bit carried away, and there's not enough information provided to determine the actual amount. It's like going to a magic show and expecting real magic. But hey, laughter is the best medicine, right?
To determine the number of grams of Ni electroplated, you will need to use Faraday's law of electrolysis. The formula for this is:
Mass of substance (in grams) = (Current (amperes) x Time (seconds) x Molar mass (grams/mole)) / (Faraday's constant x Number of electrons transferred)
Given that the current is 7.2 amperes, the time is 90.0 minutes (which needs to be converted to seconds), and the molar mass of Ni is 58.69 grams/mole, we can substitute these values into the formula.
First, we need to convert the time from minutes to seconds. There are 60 seconds in a minute, so:
Time (seconds) = 90.0 minutes x 60 seconds/minute = 5400 seconds
Next, we need to determine the number of electrons transferred. In this case, nickel (Ni) has a charge of +2 in the NiSO4 solution. So, the number of electrons transferred for one mole of Ni is 2.
Plugging all the values into the formula:
Mass of Ni = (7.2 A x 5400 s x 58.69 g/mol) / (Faraday's constant x 2)
Faraday's constant, F, is equal to 96500 C/mol.
Mass of Ni = (7.2 A x 5400 s x 58.69 g/mol) / (96500 C/mol x 2)
By solving this equation, you will get the answer for the mass of Ni electroplated.