In some industrial processes, sodium chromate is added to water coolants.
When the coolant is drained, the chromate ions can be removed through an
electrolysis process that uses an iron anode. The products of the electrolysis
are aqueous iron(II) ions and solid chromium(III) hydroxide, a recoverable
pollutant. The half-reaction involving the chromate ion is
CrO42–(aq) + 4 H2O (l) + 3 e –>>>Cr(OH)3(s) + 5 OH –(aq)
a) I figured out, thanks!
b) A current of 3.00 A is applied for 48.0 h to a cell containing a 400 g iron
anode. What is the final mass of the iron anode?
My work:
ItA/QV=m
(3)(172800)(55.85)/(9.65x10^4)(6)=50g
My new answer is 50grams. Is this right?
c) Suggest an alternative anode material that would last longer than iron. Support
your answer with relevant calculations and explanations.
My answer:
For this I got that aluminum was the answer because it had a greater electrons in half-reaction. (3V). First, is this right, and second how would I show it by calculation?
Thanks sooo much!
b) A current of 3.00 A is applied for 48.0 h to a cell containing a 400 g iron
anode. What is the final mass of the iron anode?
My work:
ItA/QV=m
(3)(172800)(55.85)/(9.65x10^4)(6)=50g
My new answer is 50grams. Is this right?
This looks ok except that isn't the 6e you used for 3 mols Fe to go into solution. (Alternatively, the half reaction showed 2e.) Second, is the 50 g what is left or what went into solution and the final mass of the iron electrode would then be 400 - 50 g (of course the 50 g will change to 150g if you decide you should use 2e instead of 6e). Check me out on this??
c) Suggest an alternative anode material that would last longer than iron. Support
your answer with relevant calculations and explanations.
My answer:
For this I got that aluminum was the answer because it had a greater electrons in half-reaction. (3V).
First, is this right, and second how would I show it by calculation?
I'm not exactly sure what is meant by this question but your answer sounds ok. You can run the same calculation by starting with 400 g Al (instead of Fe) and going through the exact calculation you did for Fe but substitute 3e for the change in electrons and substitute 27 for the molar mass instead of the 55.85. Make the calculation and see how things turn out. Remember to subtract the final anwer you obtain from 400 to find how much is left.Again, check me out on this.
b) I apologize for the mistake I made in my initial response. Let's recalculate the final mass of the iron anode using the correct values:
Given:
Current (I) = 3.00 A
Time (t) = 48.0 h
Mass of iron anode (mA) = 400 g
The formula you used, ItA/QV = m, is correct where:
I = current (A)
t = time (s)
A = molar mass of the substance (g/mol)
Q = charge (C)
V = molar volume of the substance (L/mol)
m = final mass of the substance (g)
We can break down the calculation into several steps:
1. Calculate the number of moles of iron (Fe) using the molar mass of iron (55.85 g/mol):
moles of Fe = mass of Fe / molar mass of Fe
moles of Fe = 400 g / 55.85 g/mol = 7.16 mol
2. Calculate the charge (Q) using the formula:
Q = It
Q = (3.00 A) * (48.0 h) * (3600 s/h) = 518,400 C
3. Calculate the molar volume of iron (V) using the molar volume of iron (7.092 cm3/mol or 0.007092 L/mol):
V = (mass of Fe / molar mass of Fe) * (1 / molar volume of Fe)
V = (400 g / 55.85 g/mol) * (1 / 0.007092 L/mol) = 2,524 mol/L
4. Substitute the values into the equation ItA/QV = m to find the final mass of the iron anode:
m = (ItA) / (QV)
m = (3.00 A * 518,400 C * 7.16 mol) / (518,400 C * 2,524 mol/L) = 30.4 g
Therefore, the final mass of the iron anode is 30.4 grams.
c) Your answer suggesting aluminum as an alternative anode material is correct. Aluminum has a greater number of electrons involved in its half-reaction (3 e-) compared to iron (2 e-). This indicates that aluminum can capture more electrons and undergo the redox reaction for a longer duration before getting consumed.
To show this by calculation, please follow these steps:
1. Repeat the calculations outlined in part b) with the necessary modifications:
- Calculate the number of moles of aluminum using the molar mass of aluminum (27 g/mol).
- Calculate the charge (Q) using the formula: Q = It
- Calculate the molar volume of aluminum using the molar volume of aluminum (10.00 cm3/mol or 0.01000 L/mol).
- Substitute the values into the equation ItA/QV = m to find the final mass of the aluminum anode.
2. Once you obtain the final mass of the aluminum anode, compare it to the results of the iron anode calculation. If the final mass of the aluminum anode is higher than that of the iron anode, it indicates that the aluminum anode lasts longer.
By following these steps, you can support your answer with relevant calculations and explanations.