In a dry cell, MnO2 occurs in a semisolid electrolyte paste and is reduced at the cathode. The MnO2 used in dry cells can itself be produced by an electrochemical process of which one half-reaction is shown below.
Mn2+(aq) + 2 H2O(l) MnO2(s) + 4 H+(aq) + 2 e -
If a current of 30.5 A is used, how many hours are needed to produce 1.10 kg of MnO2?
I'm sure I worked this earlier. Let me know if there is something you don't understand about the process.
To solve this problem, we can use the principles of electrochemistry and stoichiometry.
We are given that a current of 30.5 A (amperes) is used in an electrochemical process to produce MnO2. Our goal is to determine the time required to produce 1.10 kg of MnO2.
First, let's analyze the given half-reaction:
Mn2+(aq) + 2 H2O(l) -> MnO2(s) + 4 H+(aq) + 2 e-
From the balanced half-reaction, we can see that for every 2 moles of electrons (2 e-) involved in the reduction of Mn2+ to MnO2, 1 mole of MnO2 is produced.
To find the amount of MnO2 produced, we need to calculate the number of moles. We can use the molar mass of MnO2 to convert the mass given (1.10 kg) to moles.
Molar mass of MnO2 = atomic mass of Mn + (2 x atomic mass of O)
= 54.938045 g/mol + (2 x 15.999 g/mol)
= 54.938045 g/mol + 31.998 g/mol
= 86.936045 g/mol
To convert mass to moles, we can use the formula:
moles = mass / molar mass
moles of MnO2 = 1.10 kg / (86.936045 g/mol)
= 12.64 mol
Now, let's consider the Faraday's Law of Electrolysis, which states that the amount of substance produced or consumed in an electrolytic reaction is directly proportional to the quantity of electricity (electric charge) passed through the system.
We can use the equation:
moles = (coulombs / Faraday's constant) * (1 / n)
Where:
- moles is the amount of substance produced or consumed
- coulombs is the electric charge (current * time)
- Faraday's constant is the charge of 1 mole of electrons (F = 96,485 C/mol e-)
- n is the number of moles of electrons involved in the reaction (2 moles of electrons in this case)
Rearranging the equation, we can solve for time (in seconds):
time = (moles * Faraday's constant * n) / coulombs
Let's substitute the known values into the equation:
time = (12.64 mol * 96,485 C/mol * 2) / 30.5 A
Now, let's perform the calculation:
time = 779,056 s
To convert the time to hours, we divide by the number of seconds in an hour:
time = 779,056 s / 3600 s/h
≈ 216.4 hours
Therefore, it would take approximately 216.4 hours to produce 1.10 kg of MnO2 with a current of 30.5 A.