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### Topic: redox (equilibrium) problem  (Read 9534 times)

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#### escapeartistq

• Guest ##### redox (equilibrium) problem
« on: May 21, 2005, 04:20:53 PM »
Given these redox equations:

2Al(s) + 3Cu2+(aq) <=> 2Al3+(aq) + 3Cu(s)    (1)

Al(s) + 3Ag+(aq) <=> Al3+(aq) + 3Ag(s)    (2)

and knowing that K2 > K1 and that the initial concentrations of Cu2+ and Ag+ are the same (let's call the initial concentration c), which reaction results in a higher concentration of Al3+ in the equilibrium?

I tried this:

Equilibrium (1):
x is the variation of the concentration of Cu2+ until the equilibrium is achieved

2Al(s) + 3Cu2+(aq) <=> 2Al3+(aq) + 3Cu(s)
c(initial)/M                   c                   0
variation/M                -x                   +2/3 * x
c(equil)/M                  c-x                 2/3 * x

Equilibrium (2):
y is the variation of the concentration of Ag+ until the equilibrium is achieved

Al(s) + 3Ag+(aq) <=> Al3+(aq) + 3Ag(s)
c(initial)/M                   c                 0
variation/M                -y                 +1/3 * y
c(equil)/M                  c-y               1/3 * y

K1 = (2/3 * x)2/(c-x)3

K2 = (1/3 * y)/(c-y)3

Now given that K2 > K1 I don't know how to show that reaction (2) results in a higher concentration of Al3+ in the equilibrium, which is the answer to the exercise.

#### GCT

• Guest ##### Re:redox (equilibrium) problem
« Reply #1 on: May 21, 2005, 06:59:47 PM »
Quote

K1 = (2/3 * x)2/(c-x)3

K2 = (1/3 * y)/(c-y)3

Now given that K2 > K1 I don't know how to show that reaction (2) results in a higher concentration of Al3+ in the equilibrium, which is the answer to the exercise.

1...[Al3+]=(3/2)sqrt(K1(c-x)^3)
2....[Al3+]=3(K2(c-y)^3)

Let c be >>> y,x

1...[Al3+]=(3/2)sqrt(K1(c)^3)
2....[Al3+]=3(K2(c)^3)

seems to be #2

#### Borek ##### Re:redox (equilibrium) problem
« Reply #2 on: May 21, 2005, 07:33:04 PM »
Let c be >>> y,x

Seems to me this assumption is wrong - both reactions will have equilibrium shifted far to the right. Compare half potentials for Al3+/Al, Cu2+/Cu and Ag+/Ag.
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#### escapeartistq

• Guest ##### Re:redox (equilibrium) problem
« Reply #3 on: May 21, 2005, 11:43:57 PM »
Quote
Seems to me this assumption is wrong - both reactions will have equilibrium shifted far to the right. Compare half potentials for Al3+/Al, Cu2+/Cu and Ag+/Ag.

There should be some way to show that the expression for the Al3+ concentration (2=) is greater than (1).  I wonder what mathematical tools chemists use to compare such expressions (derivatives? limits?) because in this case it is not obvious just by simple inspection.

#### Borek ##### Re:redox (equilibrium) problem
« Reply #4 on: May 22, 2005, 05:16:59 AM »
There should be some way to show that the expression for the Al3+ concentration (2=) is greater than (1).  I wonder what mathematical tools chemists use to compare such expressions (derivatives? limits?) because in this case it is not obvious just by simple inspection.

The problem is - the question is IMHO stupid.

Compare half potentials of Al, Cu and Ag. You don't have to remember them exactly to know, that the difference will be huge (to be precise - 2V for Al/Cu and 2.45 for Al/Ag). It means that K2 > K1, however, both K1 and K2 are so large, that you may treat the equilibrium as completely shifted to the products side. If so, only the stochiometry has any meaning and the information about K1 and K2 is not needed for anything.

Now, if you forget about the electronegativities, and you treat the problem just as you are told, you end with 3-rd degree equations. I have a gut feeling that even if you try to solve the problem mathematically you will not have a definitive answer, as the curves describing final concentrations vs K values will cross somewhere - and depending on the K1 and K2 values either first reaction or second will be giving larger Al(3+) concentration. In other words - my intuition tells me that without exact K1 and K2 values question can't be solved. But that's only intuition.
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#### GCT

• Guest ##### Re:redox (equilibrium) problem
« Reply #5 on: May 23, 2005, 12:40:43 PM »
Perhaps this will help,

suppose that K2=K1

K2/K1=[(1/3 * y)/(c-y)3]/[(2/3 * x)2/(c-x)3]=1

Thus

(1/3)y/(c-y)^3=(2/3)x^2/(c-x)^3

(c-x)^3/(c-y)^3=(2/3)x^2/(1/3)y=2x^2/y

call the left hand side "C", that is (c-x)^3/(c-y)^3=C

y=(2/C)x^2

what you may be able to do is to plug this back in for K2 and perform some analysis and deductions from there

« Last Edit: May 23, 2005, 12:41:36 PM by GCT »

#### GCT

• Guest ##### Re:redox (equilibrium) problem
« Reply #6 on: May 23, 2005, 03:31:48 PM »
alright, so continuing from the previous post where we know the value of C

solving for K2 yields the equation

(2/3)x^2/(c-x)^3=K2, it's completely in terms of x

so [Al3+]=(2/3)x^2 for equation 2 and (4/9)x^2 for number 1.

the answer is equation number 2.  This situation is for where K2=K1, this assumption was made in the previous post.  Now if K2>K1, it would increasingly favor equation 2.  Try solving it out for yourself, the (c-y)^3 cancels out fortunately, to yield an equation completely in terms of x.

There may have been a more simpler way to solve it, but this is quite simple enough for me.  Also, are you sure that your OP had the exact form of the question?  The question seems quite useless.
« Last Edit: May 23, 2005, 03:33:00 PM by GCT »

#### Borek ##### Re:redox (equilibrium) problem
« Reply #7 on: May 23, 2005, 04:50:16 PM »
so [Al3+]=(2/3)x^2 for equation 2 and (4/9)x^2 for number 1.

Looking at the reaction equation and ICE table it seems that [Al3+] = 2/3 x in the first case, not (2/3)x^2. There is some contradiction here. Unless x = 1.

Besides, C is not a constant - it is function of x and y.
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#### GCT

• Guest ##### Re:redox (equilibrium) problem
« Reply #8 on: May 23, 2005, 05:24:33 PM »
Quote
Looking at the reaction equation and ICE table it seems that [Al3+] = 2/3 x in the first case, not (2/3)x^2. There is some contradiction here. Unless x = 1.

Besides, C is not a constant - it is function of x and y.

what???  In the first case it's ((2/3)x)^2, which turns out to be 4/9x^2, this is for the first equation.

C was integrated into K2 to solve it in terms of x, and yes, all of the x and y attributes were taken care of.  The corresponding relative aluminum concentration turned out to be 2/3x^2.

Read the WHOLE post, and perform the calculations before you blatantly answer to my posts.

#### Borek ##### Re:redox (equilibrium) problem
« Reply #9 on: May 23, 2005, 05:50:27 PM »
what???  In the first case it's ((2/3)x)^2, which turns out to be 4/9x^2, this is for the first equation.

I am referring to the very first post of whole thread - for Al/Cu reaction Al3+ concentration in equilibrium is 2/3 x. This contradicts your result for Al3+ concentration.
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#### GCT

• Guest ##### Re:redox (equilibrium) problem
« Reply #10 on: May 23, 2005, 06:07:29 PM »
ahhh, I see, you're absolutely right.  That was stupid of me (bangs head on the wall).

so [Al3+] for the first equation is (2/3)x, while for the second, it is (4/9)x^2.

I'll be rechecking the calculations to see if there are any more brain farts.