## test 2 #6 steps

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

Christineg1G
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Joined: Fri Aug 09, 2019 12:15 am
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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
Posts: 102
Joined: Fri Aug 02, 2019 12:15 am

### 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
Posts: 115
Joined: Fri Aug 09, 2019 12:15 am
Been upvoted: 1 time

### 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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Joined: Sat Aug 17, 2019 12:16 am
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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
Been upvoted: 1 time

### 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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## test 2 #6 steps

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

Christineg1G
Posts: 115
Joined: Fri Aug 09, 2019 12:15 am
Been upvoted: 1 time

### 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!
Attachments

Angus Wu_4G
Posts: 102
Joined: Fri Aug 02, 2019 12:15 am

### 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
Posts: 115
Joined: Fri Aug 09, 2019 12:15 am
Been upvoted: 1 time

### 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
Posts: 106
Joined: Sat Aug 17, 2019 12:16 am
Been upvoted: 1 time

### 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
Been upvoted: 1 time

### 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!

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

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The text passage is a series of forum posts where a student is asking for help with the mathematical steps to solve a chemistry problem, and other students are providing guidance and explanations.

TYPE: SINGLE_QUESTION_WITH_ANSWER