During the discharge of an alkaline battery, 4.50 g of Zn is consumed at the anode of the battery.
a) What mass of MnO2 is reduced at the cathode during this discharge?
b) How many Coulombs of electrical charge are transferred from Zn to MnO2?
I obtained this reaction from Wikipedia. Compare with the reaction in your text/notes. If not the same you will need to adjust my post.
Zn(s) + 2MnO2(s) is in equilibrium with ZnO(s) + Mn2O3(s) [e° = 1.43 V]
molsZn = grams Zn/atomic mass Zn
Using the coefficients in the balanced equation, convert mols Zn to mols MnO2.
Now convert mols MnO2 to g. g = mols x molar mass.
#2. mols Zn x 6.02E23 atoms/mol x 2e/atom x 1.6E-19 C/electron = ?
a) Well, to answer that, we need to see the battery shrink to fit in the palm of your hand! Just kidding! Calculating the mass of MnO2 reduced at the cathode is a piece of cake.
First, we need to know the molar mass of Zn, which is approximately 65.38 g/mol. Since 4.50 g of Zn is consumed, we can use that to find the moles of Zn consumed (4.50 g / 65.38 g/mol ≈ 0.069 moles).
Now, let's take a look at the balanced chemical equation for the reaction that occurs at the cathode during the discharge of the battery:
Zn + 2MnO2 → ZnO + Mn2O3
From the equation, we can see that 1 mol of Zn reacts with 2 mol of MnO2. So, since we know the moles of Zn consumed, the moles of MnO2 reduced will be half that amount (0.069 moles / 2 = 0.0345 moles).
To find the mass of MnO2 reduced, we multiply the moles of MnO2 by its molar mass, which is approximately 86.94 g/mol. So, the mass of MnO2 reduced is roughly 0.0345 moles * 86.94 g/mol = 3.00 g (approximately).
b) Ah, electrical charge! Sparkling science! To find the number of coulombs of electrical charge transferred from Zn to MnO2, we can use Faraday's constant, which is approximately 96,485 C/mol.
Since we know the moles of Zn consumed (0.069 moles), we can multiply it by Faraday's constant to find the total charge transferred (0.069 moles * 96,485 C/mol ≈ 6,658 C).
So, approximately 6,658 Coulombs of electrical charge are transferred from Zn to MnO2 during the battery discharge.
To determine the mass of MnO2 reduced at the cathode, we'll first need to calculate the moles of Zn consumed at the anode.
a) Calculate the moles of Zn consumed at the anode:
- The molar mass of Zn is 65.38 g/mol.
- Use the given mass of Zn (4.50 g) and divide it by its molar mass to find the number of moles.
4.50 g Zn / 65.38 g/mol = 0.0688 mol Zn
Next, we'll use the balanced chemical equation of the battery reaction to find the moles of MnO2 reduced at the cathode.
Zn + 2MnO2 -> ZnO + 2Mn
According to the equation, 1 mole of Zn reacts with 2 moles of MnO2. Hence, the number of moles of MnO2 reduced will be twice the moles of Zn consumed.
Number of moles of MnO2 reduced = 2 * 0.0688 mol Zn = 0.1376 mol MnO2
To find the mass of MnO2 reduced at the cathode, multiply the number of moles by its molar mass.
- The molar mass of MnO2 is 86.94 g/mol.
- Multiply the number of moles (0.1376 mol) by the molar mass of MnO2.
0.1376 mol MnO2 * 86.94 g/mol = 11.98 g
Therefore, the mass of MnO2 reduced at the cathode is 11.98 g.
b) To calculate the number of Coulombs of electrical charge transferred from Zn to MnO2, we'll use Faraday's law.
According to Faraday's law of electrolysis:
1 mole of electrons is equal to 1 Faraday, which is equal to 96,485 Coulombs.
From the balanced chemical equation, we know that 1 mole of Zn consumes 2 moles of electrons and 2 moles of MnO2 consumes 2 moles of electrons. Hence, the number of moles of electrons transferred will be equal to the moles of Zn consumed.
Number of moles of electrons transferred = 0.0688 mol Zn
Finally, we'll calculate the number of Coulombs of electrical charge transferred.
Number of Coulombs = Number of moles of electrons transferred * 96,485 Coulombs/mol
Number of Coulombs = 0.0688 mol Zn * 96,485 Coulombs/mol = 6,635 Coulombs
Therefore, the number of Coulombs of electrical charge transferred from Zn to MnO2 is 6,635 Coulombs.
To determine the mass of MnO2 reduced at the cathode during the discharge, we need to use the stoichiometry of the reaction occurring at the cathode.
The balanced half-reaction at the cathode during the discharge of an alkaline battery is as follows:
2MnO2 + 2H2O + 2e^- -> Mn2O3 + 4OH^-
From this reaction, we can see that 2 moles of MnO2 are reduced per 2 moles of electrons transferred. Therefore, the ratio is 1:1 between the moles of MnO2 reduced and moles of electrons transferred.
To find the moles of Zn consumed at the anode, we can use the molar mass of Zn, which is 65.38 g/mol. The molar mass allows us to convert grams of Zn to moles by dividing the mass of Zn by its molar mass:
moles of Zn = mass of Zn / molar mass of Zn
moles of Zn = 4.50 g / 65.38 g/mol
moles of Zn = 0.0688 mol
Since the ratio of moles of Zn consumed to moles of electrons transferred is also 1:1 (by definition of Faraday's law), we can conclude that the number of moles of electrons transferred is also 0.0688 mol.
Therefore, the mass of MnO2 reduced at the cathode can be calculated using the molar mass of MnO2, which is 86.94 g/mol:
mass of MnO2 = moles of MnO2 reduced x molar mass of MnO2
mass of MnO2 = 0.0688 mol x 86.94 g/mol
mass of MnO2 = 5.98 g
a) The mass of MnO2 reduced at the cathode during this discharge is approximately 5.98 g.
b) To determine the number of Coulombs transferred from Zn to MnO2, we can use Faraday's constant (F = 96,485 C/mol) which represents the charge of 1 mole of electrons. Since we know that 1 mole of electrons is transferred for every 0.0688 moles of Zn, we can calculate the number of Coulombs transferred:
Coulombs transferred = moles of electrons transferred x Faraday's constant
Coulombs transferred = 0.0688 mol x 96,485 C/mol
b) The number of Coulombs of electrical charge transferred from Zn to MnO2 is approximately 6,643 C.