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
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
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.

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