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 33.5 A is used, how many hours are needed to produce 1.60 kg of MnO2?
At which electrode is the MnO2 formed?
cathode
anode
96,485 coulombs will form approx 87/2 = approx 44 g MnO2. These are estimates and you need to redo them with more accurate numbers.
How many coulombs do you need to produce 1000 g? That's approx 96,485 x (44/1000) = ? = Q
Then Q = amperes x seconds. Substitute and solve for seconds and convert to hours.
To find the number of hours needed to produce 1.60 kg of MnO2, we need to consider the equation for the half-reaction:
Mn2+(aq) + 2 H2O(l) → MnO2(s) + 4 H+(aq) + 2 e -
From the equation, we can see that it takes 2 moles of electrons (2e-) to produce 1 mole of MnO2. We need to find the number of moles of MnO2 in 1.60 kg.
First, we convert the mass of MnO2 from kg to grams:
1.60 kg = 1600 grams
Next, we calculate the number of moles of MnO2:
Molar mass of MnO2 = 86.94 g/mol
Number of moles = Mass/Molar mass = 1600/86.94 = 18.39 moles
Now, we can calculate the number of moles of electrons required to produce 18.39 moles of MnO2:
1 mole of MnO2 → 2 moles of electrons
18.39 moles of MnO2 → X moles of electrons
X = (18.39 moles of MnO2) * (2 moles of electrons/1 mole of MnO2)
X = 36.78 moles of electrons
We know that current (I) is equal to the flow of charge (Q) per unit time. In this case, the flow of charge is given by the number of moles of electrons (n) multiplied by the Faraday constant (F).
I = n * F
Rearranging the equation, we can solve for time (t) in hours:
t = n * F / I
Where:
- n is the number of moles of electrons (36.78 moles)
- F is the Faraday constant (96485 C/mol)
- I is the current (33.5 A)
Plugging in the values, we get:
t = (36.78 moles * 96485 C/mol) / 33.5 A
Calculating this expression will give us the time required in hours to produce 1.60 kg of MnO2.