## 14.101 6th edition

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

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### 14.101 6th edition

14.101 In a neuron (a nerve cell), the concentration of K ions inside the cell is about 20–30 times as great as that outside. What potential difference between the inside and the outside of the cell would you expect to measure if the difference is due only to the imbalance of potassium ions?

Does anyone know how to solve this?

Tam To 1B
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Joined: Fri Sep 28, 2018 12:25 am

### Re: 14.101 6th edition

For this problem you will use the equation E = Ecell - (RT/nF)ln[K]. In this problem, you would look at it as E = Ecell - (RT/nF)ln[(Kout)/(Kin)]
Because both half cells are identical, you should expect Ecell = 0.
In order to find the range, you have to first calculate how E would be when ions are 30 times greater inside than outside. So you would have:
E = 0 - (RT/(1)F)ln[1/30] = +0.09 V
You do the same thing but plug in 20 instead for the K in and get +0.08 V.

Now you know the potential difference would be from +0.08 V to +0.09 V.

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