Author Topic: Blakhitman's chem questions thread! (Read 11299 times) Tweet Share

Blakhitman · « **Reply #60 on:** September 01, 2010, 09:58:07 pm »

An Electrolytic cell contains 500 mL of 0.0450 M AgNO3 solution, and a current of 0.180 A is passed through for 1.5 hours. What is the final concentration?

Need to know if I'm doing this right...Thanks!

the essence · « **Reply #61 on:** September 01, 2010, 10:20:19 pm »

n(e)=It/F

thus, n(e)=(0.180)(5400)/96500
n(e)=0.0101 mol
assuming that all Ag+(aq) is reduced to Ag(s)

n(Ag(s))=n(e) [molar ratios are the same]
= 0.0101 mol
====================

Initially we have (0.500L*0.0450M)=0.0225 mol of AgNO3 solution

The reduction of Ag+ shows us that we have used up 0.0151 mol of AgNO3(aq)

so n(AgNO3)remaining=n(AgNO3)initial-n(AgNO3)used=0.0225mol-0.0101mol=0.00124mol

so c(AgNO3)=0.00124mol/.500L=0.0248M

I think I stuffed up somewhere, feel free to point it out

edit:corrected the time

Blakhitman · « **Reply #62 on:** September 01, 2010, 10:23:25 pm »

YES I did the exact same thing

.

except I think you used wrong time, it should be 5400 sec not 8100?

Thank you so much!

the essence · « **Reply #63 on:** September 01, 2010, 11:00:54 pm »

Yep, i'll change the time

Still i'm not too sure, what is the other electrode in the electrolytic cell?

Blakhitman · « **Reply #64 on:** September 11, 2010, 03:12:53 pm »

Write the half equations for the Oxidation and reduction process in the fermentation of glucose.

98.40_for_sure · « **Reply #65 on:** September 11, 2010, 03:22:01 pm »

$C_{6}H_{12}O_{6}+12H^{+}+12e^{-}-> 3C_{2}H_{6}O + 3H_{2}O$
$C_{6}H_{12}O_{6}+6H_{2}O ->6CO_{2}+24H^{+}+24e^{-}$

If you consider $C_{6}H_{12}O_{6}$ ; the oxidation numbers of C H and O respectively are 0, +1, -2
For $C_{2}H_{6}O$ ; the oxidn numbers of C H and O are -2, +4, -2
For $6CO_{2}$ ; the oxidn numbers of C and O respectively are +4, -2
So construct two equations based on those numbers and balance using waters + oxygens

I apologize if i am incorrect

Blakhitman · « **Reply #66 on:** September 11, 2010, 05:00:02 pm »

But shouldn't the addition of the two give the equation for the fermentation of glucose?

Which is, C6H12O6 → 2C2H5OH + 2CO2.

98.40_for_sure · « **Reply #67 on:** September 11, 2010, 05:10:07 pm »

Mao · « **Reply #68 on:** September 12, 2010, 12:48:27 am »

Quote from: 99.95_for_sure on September 11, 2010, 05:10:07 pm

$2C_{3}H_{6}O_{3}+12H^{+}+12e^{-} \to 3C_{2}H_{6}O+3H_{2}O$

$C_{3}H_{6}O_{3}+3H_{2}O\to 3CO_{2}+12H^{+}+12e^{-}$

$3C_{3}H_{6}O_{3}\to 3CO_{2}+3C_{2}H_{6}O$

$C_{3}H_{6}O_{3}\to CO_{2}+C_{2}H_{6}O$

$C_{6}H_{12}O_{6}\to 2CO_{2}+2C_{2}H_{6}O$

What are you doing there? Why are you using C3H6O3?

$2C_{6}H_{12}O_{6}+24H^{+}+24e^{-}-> 6C_{2}H_{6}O + 6H_{2}O$
$C_{6}H_{12}O_{6}+6H_{2}O ->6CO_{2}+24H^{+}+24e^{-}$

Adding them together will give

$3C_{6}H_{12}O_{6} \to 6CO_2 + 6 C_{2}H_{6}O$ , divide by three will be what you want.

98.40_for_sure · « **Reply #69 on:** September 12, 2010, 10:50:31 am »

Lol can't you split the glucose into
$C_{3}H_{6}O_{3}+C_{3}H_{6}O_{3}$

Mao · « **Reply #70 on:** September 12, 2010, 12:29:39 pm »

Glucose:

Triose (D-Glyceraldehyde):

No, you cannot split organic molecules just because.

Blakhitman · « **Reply #71 on:** September 12, 2010, 01:22:18 pm »

Quote from: Mao on September 12, 2010, 12:48:27 am

Quote from: 99.95_for_sure on September 11, 2010, 05:10:07 pm
$2C_{3}H_{6}O_{3}+12H^{+}+12e^{-} \to 3C_{2}H_{6}O+3H_{2}O$

$C_{3}H_{6}O_{3}+3H_{2}O\to 3CO_{2}+12H^{+}+12e^{-}$

$3C_{3}H_{6}O_{3}\to 3CO_{2}+3C_{2}H_{6}O$

$C_{3}H_{6}O_{3}\to CO_{2}+C_{2}H_{6}O$

$C_{6}H_{12}O_{6}\to 2CO_{2}+2C_{2}H_{6}O$

What are you doing there? Why are you using C3H6O3?

$2C_{6}H_{12}O_{6}+24H^{+}+24e^{-}-> 6C_{2}H_{6}O + 6H_{2}O$
$C_{6}H_{12}O_{6}+6H_{2}O ->6CO_{2}+24H^{+}+24e^{-}$

Adding them together will give

$3C_{6}H_{12}O_{6} \to 6CO_2 + 6 C_{2}H_{6}O$ , divide by three will be what you want.

Oh man, and I actually tried adding it, I fail because instead of adding the glucose I subtracted for some unkown reason.

Thanks so much man.

m@tty · « **Reply #72 on:** September 12, 2010, 02:48:04 pm »

Quote from: 99.95_for_sure on September 12, 2010, 10:50:31 am

Lol can't you split the glucose into
$C_{3}H_{6}O_{3}+C_{3}H_{6}O_{3}$

Lol.. Nice try...

Blakhitman · « **Reply #73 on:** October 12, 2010, 04:55:32 pm »

What do you guys say the answer is?

Studyinghard · « **Reply #74 on:** October 12, 2010, 04:59:57 pm »

I would say B. whats the answer?

Search the forums now!

We have moved!

Author Topic: Blakhitman's chem questions thread! (Read 11299 times) Tweet Share

Blakhitman

Re: Blakhitman's chem questions thread!

the essence

Re: Blakhitman's chem questions thread!

Blakhitman

Re: Blakhitman's chem questions thread!

the essence

Re: Blakhitman's chem questions thread!

Blakhitman

Re: Blakhitman's chem questions thread!

98.40_for_sure

Re: Blakhitman's chem questions thread!

Blakhitman

Re: Blakhitman's chem questions thread!

98.40_for_sure

Re: Blakhitman's chem questions thread!

Mao

Re: Blakhitman's chem questions thread!

98.40_for_sure

Re: Blakhitman's chem questions thread!

Mao

Re: Blakhitman's chem questions thread!

Blakhitman

Re: Blakhitman's chem questions thread!

m@tty

Re: Blakhitman's chem questions thread!

Blakhitman

Re: Blakhitman's chem questions thread!

Studyinghard

Re: Blakhitman's chem questions thread!

Recent Posts

User:
Password: