June 18, 2021, 11:49:16 PM
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### Topic: Calculate ε  (Read 195 times)

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#### Judy

• Regular Member
•   • Posts: 34
• Mole Snacks: +0/-0 ##### Calculate ε
« on: June 08, 2021, 01:10:46 AM »
Calculate ε at 25°C for the cell shown below, given the following data:

1.0 x 10^-3M Ni2+ 1.0M HCl
1.0M HCl AgCl(s)

Ksp for AgCl = 1.6 x 10^-10
A) 0.83V
B) 0.54V
C) 1.01V
D) 2.98V
E) cannot be determined from the data given

ANS: B

The answer I calculated is close to A, but the given answer is B.
Ni + 2Ag → Ni2+  + 2Ag
ε = 1.05 - 0.0596*0.5*log[10^-3/(√1.6x10^-10)^2]
= 0.85

#### Borek ##### Re: Calculate ε
« Reply #1 on: June 08, 2021, 03:19:10 AM »
(√1.6x10^-10)^2

This doesn't make much sense, suggesting something is wrong with the equation.
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#### Judy

• Regular Member
•   • Posts: 34
• Mole Snacks: +0/-0 ##### Re: Calculate ε
« Reply #2 on: June 08, 2021, 11:17:24 AM »
Since [Ag+] is (1.6e^-10)^1/2 derived from the Ksp value and is twice of [Ni2+] based upon my chemical reaction (Ni + 2Ag+ -> Ni2+ + 2Ag), the [Ag+] should be 1.6e^-10 in the denominator within the log(). So, I don't understand why it doesn't make sense.

Is there any other hint? Thank you.

#### mjc123

• Chemist
• Sr. Member
• • Posts: 1917
• Mole Snacks: +275/-12 ##### Re: Calculate ε
« Reply #3 on: June 08, 2021, 05:41:12 PM »
[Ag+] is not equal to the square root of the solubility product when you have 1M HCl present.

#### Judy

• Regular Member
•   • Posts: 34
• Mole Snacks: +0/-0 ##### Re: Calculate ε
« Reply #4 on: June 08, 2021, 10:54:52 PM »
I got it. Thanks so much.