## test 2 #6 steps

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Christineg1G
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### test 2 #6 steps

Can someone please explain the math steps involved to get the correct answer for this problem? I understand the set up, just not the steps you need to take to get the right answer of 0.42M. Thank you!
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Angus Wu_4G
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### Re: test 2 #6 steps

I think the mistake you did was using a wrong value in the equation, instead of using 0.257/2, you should've used 0.0592/n, where n is the number of mole of electrons.

Christineg1G
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### Re: test 2 #6 steps

Angus Wu_4G wrote:I think the mistake you did was using a wrong value in the equation, instead of using 0.257/2, you should've used 0.0592/n, where n is the number of mole of electrons.

That change in value shouldn't matter because you should still receive the same answer. I'm just having trouble with the calculus part of determining the "x" from the ln x/2.2.

Diana A 2L
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### Re: test 2 #6 steps

Christineg1G wrote:Can someone please explain the math steps involved to get the correct answer for this problem? I understand the set up, just not the steps you need to take to get the right answer of 0.42M. Thank you!

Well first you’d multiply both sides by -2/0.0257 ( the reciprocal of the constant on the left side of the ln) to cancel it out. Then you get ln (x/2.2)= 0.0213(-2/0.0257). Then you exponentiate the equation and you get e^(0.0213(-2/0.0257))= x/2.2 and then you multiply each side by 2.2 and you get the value for x. You can do this bc the rule is log(base a) k = x becomes a^x =k. And for ln it’s ln(base e) k= x, which becomes e^x =k. Hope that helps.

Christineg1G
Posts: 115
Joined: Fri Aug 09, 2019 12:15 am
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### Re: test 2 #6 steps

Diana A 2L wrote:
Christineg1G wrote:Can someone please explain the math steps involved to get the correct answer for this problem? I understand the set up, just not the steps you need to take to get the right answer of 0.42M. Thank you!

Well first you’d multiply both sides by -2/0.0257 ( the reciprocal of the constant on the left side of the ln) to cancel it out. Then you get ln (x/2.2)= 0.0213(-2/0.0257). Then you exponentiate the equation and you get e^(0.0213(-2/0.0257))= x/2.2 and then you multiply each side by 2.2 and you get the value for x. You can do this bc the rule is log(base a) k = x becomes a^x =k. And for ln it’s ln(base e) k= x, which becomes e^x =k. Hope that helps.

Thank you!

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