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Hey...how r u guys..thanks for helping me on my previous questions...can u help me solve two questions and check one? Thanks, all the questions are related to the info given:

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)


21. a) Provide the half-reactions and a net redox reaction for this electrochemical
process.

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 answer:
The final mass of iron anode is 700grams.


c) Suggest an alternative anode material that would last longer than iron. Support
your answer with relevant calculations and explanations.


How can the final mass of iron be greater than the initial mass? It seems to me that you are an answer moocher, fishing for answers. Do you ever show your work?


ummm...that is my answer and I just want to know how to do this. I am not that great at chem, but want to understand the questions I do. If u hate it, fine! I just want to know how to do this.


Hey, sorry anonymous, I am just real frustrated from chem cause I don't get it and I have my final on Tuesday. I shouldnt have got mad, but I did and sorry!


Here are some hints.

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)


21. a) Provide the half-reactions and a net redox reaction for this electrochemical
process.
The problem gives one half reaction in the CrO4^2- going to Cr(OH)3(s). The second half reaction is given in the preamble as the iron electrode going into solution as the ferrous ion; therefore, that half reaction is Fe(s) ==> Fe^2+(aq) + 2e
Now you want to add these two equations together. To do this, however, you must first make the electrons lost in the first reaction (3e) and the electrons gained in the iron reaction (2e) equal. You do that by multiplying the iron half reaction by 3 and the chromate half reaction by 2 (that makes six electrons lost and six electrons gained). Then add those two equations (the ones you obtain from multiplying) together.


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 answer:
The final mass of iron anode is 700grams.
Show us what you did on this. We need you to show your work in order to understand where you may have gone wrong. For example, the preample tells us that the iron electrode is going into solution as ferrous ion (that is Fe^+2). Therefore, I would expect the final mass of the iron electrode to be less than 400 grams.

c) Suggest an alternative anode material that would last longer than iron. Support
your answer with relevant calculations and explanations.
I would look for an element that takes more coulombs to convert from metal to ions. That means look for a metal that has a half reaction involving more than 2 electrons. The activity series might be the place to start.I hope this helps. Please post any follow up questions just one question per post. It is more convenient to handle that way and show any work you have done. Finally, tell us exactly what you don't understand.

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