If we electrolyze a solution of Ni²⁺(aq) to form Ni(s) and use a current of 0.15 amps for 10 minutes, how many grams of Ni(s) are produced?
To follow up on oobleck's observation, you posted about a dozen questions a few days ago in less than a minute. I posted answers to all of those although I saw no effort on your part. Today you've posted a dozen but it took you longer. I've not looked at all of them but the two I did I saw no effort on your part. Our goal on Jiskha is to HELP. This is not a homework dump site. Please go back and show us what you know on the other problems. Explain what you don't understand about the problem and where you need help and someone will gladly help you through what you don't understand. Here is how to proceed on this problem.
How many coulombs do you have? That's
C = amperes x seconds = 0.15 x 10 min x((60 s/min) = 90
It will take 96,485 coulombs to deposit 58.7/2 = 29.3 g Ni.
g Ni deposited = 29.3 g x (90/96,485) = ?
To determine the number of grams of Ni(s) produced, we need to calculate the amount of charge passed during electrolysis and then use Faraday's law of electrolysis to convert the charge to moles of Ni(s), and finally, convert moles to grams.
Step 1: Calculate the total charge passed (Q)
Charge (Q) = Current (I) × Time (t)
Given:
Current (I) = 0.15 A
Time (t) = 10 minutes = 10 × 60 = 600 seconds
Q = 0.15 A × 600 s
Q = 90 Coulombs
Step 2: Convert charge (Q) to moles of electrons (n)
1 mole of electrons = 1 Faraday (F) = 96,485 Coulombs
n = Q / F
n = 90 C / 96,485 C/mol
n ≈ 0.000931 mol
Step 3: Apply Faraday's law of electrolysis to determine the moles of Ni(s) produced.
For the balanced equation: Ni²⁺(aq) + 2e⁻ → Ni(s)
1 mole of Ni²⁺(aq) reacts with 2 moles of electrons.
Therefore, the moles of Ni(s) produced = 0.000931 mol × 1 mol Ni(s) / 2 mol electrons
Moles of Ni(s) = 0.0004655 mol
Step 4: Convert moles of Ni(s) to grams using the molar mass of nickel (Ni).
The molar mass of nickel (Ni) = 58.69 g/mol
Mass of Ni(s) = Moles of Ni(s) × Molar mass of Ni
Mass of Ni(s) = 0.0004655 mol × 58.69 g/mol
Mass of Ni(s) ≈ 0.0273 g
Therefore, approximately 0.0273 grams of Ni(s) are produced.
To calculate the mass of Ni(s) produced during the electrolysis of a solution of Ni²⁺(aq), we need to apply Faraday's laws of electrolysis.
Faraday's first law states that the mass of a substance produced by electrolysis is directly proportional to the quantity of electricity passed through it. The equation that relates mass, quantity of electricity, and the electrochemical equivalent is:
Mass (g) = (Quantity of Electricity (Coulombs) × Electrochemical Equivalent (g/C)) / Charge (Coulombs per mole)
In this case, we need to find the mass of Ni(s) produced, so our formula becomes:
Mass of Ni(s) = (Quantity of Electricity × Electrochemical Equivalent of Ni(s)) / Charge per mole of Ni²⁺(aq)
To find the electrochemical equivalent of Ni(s), we need to know the stoichiometry of the reaction. Since the electrolysis of Ni²⁺(aq) produces Ni(s) with a 2:2 ratio (2 electrons are transferred during the reaction), we can deduce that the electrochemical equivalent of Ni(s) is equal to the molar mass of Ni divided by 2:
Electrochemical Equivalent of Ni(s) = Molar mass of Ni(g/mol) / 2
Now, we can proceed with the calculations:
1. Calculate the quantity of electricity passed:
Quantity of Electricity = Current (A) × Time (s)
= 0.15 A × (10 minutes × 60 seconds/minute)
= 0.15 A × 600 s
= 90 Coulombs
2. Calculate the electrochemical equivalent of Ni(s):
Molar mass of Ni = 58.69 g/mol (atomic mass of Ni)
Electrochemical Equivalent of Ni(s) = 58.69 g/mol / 2
= 29.35 g/C
3. Calculate the number of Coulombs per mole of Ni²⁺(aq):
Ni²⁺(aq) + 2e⁻ → Ni(s) indicates that 2 moles of electrons are required to produce 1 mole of Ni(s)
Therefore, the ratio is 2 Coulombs : 1 mole of Ni²⁺(aq)
4. Finally, calculate the mass of Ni(s):
Mass of Ni(s) = (Quantity of Electricity × Electrochemical Equivalent of Ni(s)) / Charge per mole of Ni²⁺(aq)
= (90 C × 29.35 g/C) / 2 C/mole
= 1,307.5 g / 2
= 653.75 g
Therefore, approximately 653.75 grams of Ni(s) are produced.
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