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Valentin [98]
2 years ago
14

A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached

to the positive terminal of a potentiometer and the calomel electrode was attached to the negative terminal.(a) Write a half-reaction for the Cu electrode. (Use the lowest possible coefficients. Omit states-of-matter.)
(c) Calculate the cell voltage.
Chemistry
1 answer:
lara [203]2 years ago
6 0

Answer:

a)  cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

b)  0.068 V.

Explanation:

A) Cu2+ + 2e- euilibrium cu (s)

 Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-

Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

B) To calculate the cell voltage

E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+

putting values we get

 = 0.339V + (90.05916V/2)log(0.100) = 0.309V

 E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.

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Explanation:

Step 1:

A good first step for a problem like this is to write down the chemical formula and balance it.

It appears here that we have 10.5 mL of vinegar, which IS acetic acid, and 19.13 mL of 0.460 M NaOH. That will give us the following balanced chemical equation:

CH3COOH + NaOH ------> NaCH3COO + H2O

All of the constituents come out to a value of 1, conveniently.

Step 2:

Since all of our stoichiometric coefficients are one, we can use a shortcut to answer this equation. I don't know if it has a name, but I just call it the titration formula. It goes something like this:

M1 * V1 = M2 * V2

M stands for Molarity and V stands for volume. 1 and 2 being the before the reaction and after the reaction.

So, our M1 for this is going to be what the question says was used for this titration. That's 0.460M NaOH.

Our V1 is going to be the initial volume of the sample, which was 10.5 mL

Our V2 is going to be 19.13, which is the volume when we're finished.

It's clear that we don't know M2, so let's find it.

Keep in mind that it's easier to convert to liters pretty much always, so I've done that by dividing the mL values each by 1000.

Using some algebra, we can see that we now have:

0.460 M * 0.0105 L = x M * 0.01913 L

Which goes to:

\frac{0.00483mol}{0.01913L} = 0.252 M

<h3>So our M2, the molar concentration of acetic acid in this vinegar, is equal to 0.252 M. </h3>
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