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Offline alanjz

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Electrochemistry Question
« on: April 04, 2013, 09:05:50 PM »
I'm having trouble solving this problem. I just don't know where to begin. I think it uses Nernst equation but I can't seem to figure it out.

39. The standard reduction potential for H+ (aq) is 0.00 V.
What is the reduction potential for a 1×10-3 M HCl solution?
(A) 0.355 V (B) 0.178 V
(C) –0.178 V (D) –0.355 V

The answer is C.

Offline Stovn0611

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Re: Electrochemistry Question
« Reply #1 on: April 04, 2013, 11:31:11 PM »
Yes, the Nernst equation would be used to solve this problem.

The first step is to identify the reaction (reduction of HCl)

Cl is in its lowest oxidation state at -1 already so only H can be reduced from +1 to 0

2H+ + 2e- -> H2

Now use the Nernst equation

E = 0.00 V - (RT)/(nF)ln(Q)

R = 8.3145 J/mol*K (constant)
T = 298 K (standard conditions)
F = 96485 C/mol (Faraday constant)
n = number of electrons transferred
Q = equilibrium expression of reduction reaction

Offline Borek

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Re: Electrochemistry Question
« Reply #2 on: April 05, 2013, 04:34:27 AM »
Problem is, there is no mentioning of the partial pressure of H2, so question is not a straightforward as it may look.

The only way to get an answer is to assume standard hydrogen pressure (1 atm).
« Last Edit: April 05, 2013, 05:54:20 AM by Borek »
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Offline alanjz

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Re: Electrochemistry Question
« Reply #3 on: April 05, 2013, 08:19:31 AM »
would I use 273 or 298 for the temperature and why is the answer negative? I get that it's positive when I was plug into the equation.

Offline Borek

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Re: Electrochemistry Question
« Reply #4 on: April 05, 2013, 12:33:00 PM »
would I use 273 or 298 for the temperature

Make your pick - this is one of the things that are not stated in the question. I would go with 25°C (298K).

Quote
and why is the answer negative? I get that it's positive when I was plug into the equation.

Most likely you are confusing sign conventions.
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Offline alanjz

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Re: Electrochemistry Question
« Reply #5 on: April 05, 2013, 12:48:00 PM »
Could you explain a little more? I plugged in 8.314 for R, 96500 for F, 273 for T, 2 for n and 1E-6 for Q and I got a positive value for E.

Offline Borek

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Re: Electrochemistry Question
« Reply #6 on: April 06, 2013, 03:31:46 AM »
Show the exact equation you used.
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Offline alanjz

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Re: Electrochemistry Question
« Reply #7 on: April 06, 2013, 09:55:04 AM »
E = 0.00 V - (8.314)(273)/(2*96500)ln(1e-6)

Offline Borek

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Re: Electrochemistry Question
« Reply #8 on: April 06, 2013, 09:59:23 AM »
Please write it using symbols, so that it will be clear what is your Q.
« Last Edit: April 06, 2013, 05:10:37 PM by Borek »
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Offline alanjz

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Re: Electrochemistry Question
« Reply #9 on: April 06, 2013, 10:01:43 AM »
What do you mean by symbols? I just used Nernst equation and my Q was 10^-6 M because the HCl dissociates

Offline Borek

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Re: Electrochemistry Question
« Reply #10 on: April 06, 2013, 12:52:00 PM »
Symbols as opposed to values.
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Offline alanjz

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Re: Electrochemistry Question
« Reply #11 on: April 06, 2013, 01:39:56 PM »
E = 0.00 V - (RT)/(nF)ln(Q)
this? where Q is [H] [Cl]

Offline Borek

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Re: Electrochemistry Question
« Reply #12 on: April 06, 2013, 05:11:30 PM »
Write explicitely your Q. It should reflect the reaction.
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Offline alanjz

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Re: Electrochemistry Question
« Reply #13 on: April 06, 2013, 06:02:25 PM »
ah I think I got it: the equation is H+ + Cl- --> H2+Cl2 so Q would be 1/(10^-3*10^-3) which makes the final E negative. Lemme know if this is right. Thanks for your patience with me :)

Offline Borek

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Re: Electrochemistry Question
« Reply #14 on: April 07, 2013, 03:55:35 AM »
No, this is not the correct reaction. Question asks about the reduction of H+ - what is H+ reduced to?
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