stonecold's chem questions :)

[toshibaj]

How do you do the half cell equations for this...

Ag2O(s) + Zn(s) + H2O(l) --->  2Ag(s) + Zn(OH)2(s)

I have no idea whether to balance with H2O, OH-, H+, or O2- in these types of questions...

4e- + Ag2O + 2H2O --> 2Ag(s) + 4OH-
Zn(s) + 2OH- --> Zn(OH)2 + 2e-

There are acidic redox reactions and basic redox reactions. Acidic involves H+, and basic involves OH-.

Solutions say this :S

Ag2O + H2O + 2e- --> 2Ag(s) + 2OH-

I agree with stonecold's solutions. Not sure how kyzoo got that, great genius that he is...=/.

Start:

Ag2O   ---> 2Ag

add h20

Ag2O ---> 2Ag + H20

add H+

Ag2O + 2H+   -----> 2Ag +H20

because this is in acidic electrolyte (you can tell by the ZN(OH)2), add OH- for every H+:


Ag2O +2H2O  -----> 2Ag + H20 + 2OH-

cancel out H2O:

Ag2O + H2O  ----> 2Ag + 2OH-

wow that last thing where you added the OH- to both sides for each H+ was genius.

i never knew that it was possible to work these out methodically. i've always done trial and error lol.

thanks!

[kyzoo]

0.o LOL my mistake It hought there were 2 oxygens in Ag2O

And yea xD Iffets you are a genius ^.^ I thought it was trial and error as well

[iffets12345]

wow that last thing where you added the OH- to both sides for each H+ was genius.

i never knew that it was possible to work these out methodically. i've always done trial and error lol.

thanks!
[/quote]

Thank Mao, and other various chem websites .

[stonecold]

legend iffets.  i remember mao doing this but i forgot it.

thanks for showing it again.

[stonecold]

I heard we just had to be familiar, but I guess knowing wouldnt hurt.
Except, I have a sheet with 20 different kinds of fuel cells. FML

Balance the equation as you would for anything in an acidic environment (key element, water, H+, electrons)

Next, if it is an alkaline cell, add enough OH- to both sides to neutralize any H+ in the equation.
If it is a solid oxide fuell cell, add enough O2- to both sides to neutralize any H+ in the equation.

Voila, all the fuel cells learnt.

I am just bumping this.  Everyone needs to know it.  It is so simple and easy to understand.

Mao, you are awesome, and Heinemann should be paying you thousands of $$ to put simple, yet insightful explanations like this in their textbook.

[Mao]

Now I feel like a real scientist, getting recognition months after the initial publication

lol I'm kidding, I won't take any credit for it, my awesome chem teacher taught me that back in year 12.

[iffets12345]

Mao would you like me to ruin the 999 karma and make it 1k?

[Blakhitman]

Mao would you like me to ruin the 999 karma and make it 1k?

NO I JUST ADDED TO HIS KARMA TO MAKE IT 999!!!

[iffets12345]

damn it's 1k already.

[Mao]

I challenge you to make it to the next palindrome (1111 is the next one )

[iffets12345]

I shall be waiting!

[stonecold]

How do you go about this question.  You can work it out because all the wrong answers are false statements, but NEAP says something about using the E value...

[iffets12345]

How do you go about this question.  You can work it out because all the wrong answers are false statements, but NEAP says something about using the E value...

I guess if you know the E value, you will know the zinc equation's relative position to the manganese one. If it's below then it oxidises.

[stonecold]

How do you go about this question.  You can work it out because all the wrong answers are false statements, but NEAP says something about using the E value...

I guess if you know the E value, you will know the zinc equation's relative position to the manganese one. If it's below then it oxidises.

Yeah, but there is no Mn for +3 and +4 oxidation states on the ECS...

[iffets12345]

it gives you the E in the question