Chemical Forums

Chemistry Forums for Students => High School Chemistry Forum => Topic started by: alanjz on April 04, 2013, 09:05:50 PM

Title: Electrochemistry Question
Post by: alanjz 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.
Title: Re: Electrochemistry Question
Post by: Stovn0611 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
Title: Re: Electrochemistry Question
Post by: Borek 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).
Title: Re: Electrochemistry Question
Post by: alanjz 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.
Title: Re: Electrochemistry Question
Post by: Borek on April 05, 2013, 12:33:00 PM
Quote from: alanjz on April 05, 2013, 08:19:31 AM
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.
Title: Re: Electrochemistry Question
Post by: alanjz 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.
Title: Re: Electrochemistry Question
Post by: Borek on April 06, 2013, 03:31:46 AM
Show the exact equation you used.
Title: Re: Electrochemistry Question
Post by: alanjz on April 06, 2013, 09:55:04 AM
E = 0.00 V - (8.314)(273)/(2*96500)ln(1e-6)
Title: Re: Electrochemistry Question
Post by: Borek on April 06, 2013, 09:59:23 AM
Please write it using symbols, so that it will be clear what is your Q.
Title: Re: Electrochemistry Question
Post by: alanjz 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
Title: Re: Electrochemistry Question
Post by: Borek on April 06, 2013, 12:52:00 PM
Symbols as opposed to values.
Title: Re: Electrochemistry Question
Post by: alanjz on April 06, 2013, 01:39:56 PM
E = 0.00 V - (RT)/(nF)ln(Q)
this? where Q is [H] [Cl]
Title: Re: Electrochemistry Question
Post by: Borek on April 06, 2013, 05:11:30 PM
Write explicitely your Q. It should reflect the reaction.
Title: Re: Electrochemistry Question
Post by: alanjz 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 :)
Title: Re: Electrochemistry Question
Post by: Borek 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?
Title: Re: Electrochemistry Question
Post by: alanjz on April 07, 2013, 10:46:56 AM
isn't the reduction: 2H+ + 2e- --> H2?
Title: Re: Electrochemistry Question
Post by: Borek on April 07, 2013, 10:56:18 AM
Yes, that's a correct reaction now.
Title: Re: Electrochemistry Question
Post by: alanjz on April 07, 2013, 10:57:13 AM
ah i see, i didn't balance it before right?