Jul 17, 2010
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I understand the concept, but I can't seem to figure out the half reactions. Can someone please explain this to me?

Thank you in advance.

 

Jab1113

7+ Year Member
May 17, 2011
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Ann Arbor, MI
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I understand the concept, but I can't seem to figure out the half reactions. Can someone please explain this to me?

Thank you in advance.

For me personally, I like to use the "OIL RIG" instead of "LEO GER", I just remember it better...

Oxidation is loss of e-
Reduction is gain of e-

We also know that "RED CAT, AN OX", so reduction at the cathode, oxidation at the anode.

So reduction at the cathode is gain of e- so an example would be...

Al+3 + 3e- ----> Al

And oxidation at the anode is loss of e- so example would be....

2Cl ----> Cl2 + 2e-

REMEMBER: anions will go to anode. cations will go to cathode.
 

LetsGo2DSchool

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Jul 11, 2009
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Have no fears. This type of question can be answered so quickly and effortlessly. You don't even need to write out the half rxns.

It's asking for the reducing agent which means it gets oxidized. Thus, look for the reactant species that loses electrons (or the thing that will go from neutral to positive, or negative to neutral).

Thus, it has to be Al
 
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Jul 17, 2010
11
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Status
Pre-Dental
Thank you both. I guess my problems is recognizing which one is losing electron and which one is gaining. I just can't get a hang of it.

Can you please explain your technique?

For example, is my assumption correct here?

Given:
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

is this right?
Al --> Al3+ + 3e-
SO4 + 2e- --> SO4-2
 

Jab1113

7+ Year Member
May 17, 2011
224
10
Ann Arbor, MI
Status
Dental Student
Thank you both. I guess my problems is recognizing which one is losing electron and which one is gaining. I just can't get a hang of it.

Can you please explain your technique?

For example, is my assumption correct here?

Given:
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

is this right?
Al --> Al3+ + 3e-
SO4 + 2e- --> SO4-2
Remember Cations will go to the Cathode. Cathode = Reduction = Gain of e- (this will go on the left side of the equation)

Al+3 + 3e- ---> Al

Then you have Anions which will go to the Anode. Anode = Oxidation = Loss of e- (this will go on the right side of the equation)

SO4 ----> SO4-2 + 2e-

Somebody please make sure I did this right, I dont want someone to memorize the wrong information lol.
 

toothhornet88

7+ Year Member
Feb 6, 2010
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Remember Cations will go to the Cathode. Cathode = Reduction = Gain of e- (this will go on the left side of the equation)

Al+3 + 3e- ---> Al

Then you have Anions which will go to the Anode. Anode = Oxidation = Loss of e- (this will go on the right side of the equation)

SO4 ----> SO4-2 + 2e-

Somebody please make sure I did this right, I dont want someone to memorize the wrong information lol.
Your right but I wouldn't even bring Cathode/Anode in for this type of question. Just know OIL RIG
So you're looking for a compound that was oxidized (loses electrons)

2Al(s) has a zero charge and ends up with a + charge on the products side so it lost electrons AKA got oxidized... and there you go you got your answer

P.S. Whenever the question asks about a reducing/oxidizing agent always change it to oxidized/reduced compound to make it easier and not get mixed up
 
Jul 17, 2010
11
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Pre-Dental
Thank you so much guys! I'm getting them right now. Really appreciate it.

And thanks for that tip, toothhornet88. It works wonders. :D
 
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