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.

a) The half-reactions for this electrochemical process are:

1) CrO42–(aq) + 4 H2O (l) + 3 e –>>>Cr(OH)3(s) + 5 OH –(aq)
2) Fe(s) >>> Fe^2+(aq) + 2e

To obtain the net redox reaction, both half-reactions need to be balanced by multiplying them by appropriate factors so that the electrons lost in one half-reaction are equal to the electrons gained in the other half-reaction.

Multiply the first half-reaction by 2 and the second half-reaction by 3 to balance the electrons:

2CrO42–(aq) + 8 H2O (l) + 6 e –>>> 2Cr(OH)3(s) + 10 OH –(aq)
3Fe(s) >>> 3Fe^2+(aq) + 6e

Now add the balanced half-reactions together to obtain the net redox reaction:

2CrO42–(aq) + 3Fe(s) + 8 H2O (l) >>> 2Cr(OH)3(s) + 3Fe^2+(aq) + 10 OH –(aq)

b) To calculate the final mass of the iron anode, we need to use Faraday's Law of electrolysis, which states that the mass of a substance produced or consumed in an electrochemical reaction is directly proportional to the amount of electric charge passed through the cell.

The formula to calculate the mass of a substance consumed or produced is given by:
Mass = (I * t * M) / (n * F)

Where:
I = current (in amperes)
t = time (in seconds)
M = molar mass of the substance (in grams/mole)
n = number of electrons transferred in the balanced half-reaction
F = Faraday's constant (approximately 96,485 C/mol)

In this case, we are interested in the consumption of iron, so the formula becomes:
Mass of iron anode = (I * t * M_Fe) / (n_Fe * F)

Given:
Current (I) = 3.00 A
Time (t) = 48.0 h = 48.0 * 60 * 60 s = 172,800 s
Molar mass of iron (M_Fe) = 55.85 g/mol
Number of electrons transferred in the iron half-reaction (n_Fe) = 2
Faraday's constant (F) = 96,485 C/mol

Substituting these values into the formula, we can calculate the final mass of the iron anode.

Mass of iron anode = (3.00 A * 172,800 s * 55.85 g/mol) / (2 * 96,485 C/mol)

Calculating the above expression will give you the final mass of the iron anode.

c) To suggest an alternative anode material that would last longer than iron, we need to consider a metal with a half-reaction that involves more than 2 electrons. This is because the more electrons involved in the reaction, the slower the anode material will be consumed.

One way to find such a metal is by looking at the activity series of metals. Metals higher up in the activity series are typically more reactive and have half-reactions involving more electrons. For example, metals like platinum, gold, and silver have half-reactions involving multiple electrons.

Therefore, a possible alternative anode material could be platinum (Pt) or another metal higher up in the activity series. These metals would last longer because they would be consumed at a slower rate due to their half-reactions involving more electrons.

Please keep in mind that this is just a suggestion based on the information provided. It is always recommended to consult additional sources and consider other factors before selecting an alternative anode material.