Test 2


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204929947
Posts: 76
Joined: Fri Apr 06, 2018 11:03 am

Test 2

Postby 204929947 » Sun Mar 03, 2019 12:00 pm

Okay, so for Test 2, it had us put the redox molecules from increasing order, how would we do that?

Kenan Kherallah 2C
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

Re: Test 2

Postby Kenan Kherallah 2C » Sun Mar 03, 2019 12:53 pm

You would look at the reducing power which is another way of saying how likely the substance is to be oxidized. This means the more negative the E potential the better.

Henry_Phan_4L
Posts: 68
Joined: Fri Sep 28, 2018 12:24 am

Re: Test 2

Postby Henry_Phan_4L » Sun Mar 03, 2019 12:54 pm

So the one with better oxidizing power is the most positive one?

Jonathan Zhao 4H
Posts: 57
Joined: Fri Sep 28, 2018 12:26 am

Re: Test 2

Postby Jonathan Zhao 4H » Sun Mar 03, 2019 1:26 pm

Yes, correct. The more positive the E value, the stronger the oxidizing power.

shouse1f
Posts: 63
Joined: Fri Sep 28, 2018 12:18 am

Re: Test 2

Postby shouse1f » Wed Mar 06, 2019 12:16 pm

On test 2 for another part of the same question it asked which 2 species would give the largest E cell, how do you determine this?

Tatum Keichline 2B
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

Re: Test 2

Postby Tatum Keichline 2B » Wed Mar 06, 2019 12:50 pm

More negative E for reducing and more positive E for oxidizing I believe

sonalivij
Posts: 70
Joined: Fri Sep 28, 2018 12:27 am

Re: Test 2

Postby sonalivij » Wed Mar 06, 2019 2:22 pm

The half reaction with the most negative standard potential is the strongest reducing agent, and the half reaction with the most positive standard potential is the strongest oxidizing agent

Selina Bellin 2B
Posts: 62
Joined: Fri Sep 28, 2018 12:25 am

Re: Test 2

Postby Selina Bellin 2B » Wed Mar 06, 2019 9:29 pm

more positive e value corresponds to a higher oxidizing power


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