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Vitek1552 [10]
3 years ago
9

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

5.20×10?2 M and 1.40 M , respectively.
Part A I have part A answered but need help with Part B and C...below.

What is the initial cell potential?

Express your answer using two significant figures.

Ecell =
0.51

V
Part B

What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M?

Express your answer using two significant figures.

Ecell =
V

Part C

What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V ?

Enter your answers numerically separated by a comma. Express your answer using two significant figures.
Chemistry
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

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5 0
3 years ago
Calculate the percent by mass of each element in
Zepler [3.9K]
The elements present in Ammonium Nitrate are Hydrogen, Nitrogen, and Oxygen at a ratio of 4:2:3, respectively. Hydrogen weighs in at 1.008 amu, Nitrogen at 14.007, and Oxygen at 15.999. This means that the molar mass would be:

Hydrogen
4 x 1.008 = 4.032 amu

Nitrogen
2 x 14.007 = 28.014 amu

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3 x 15.999 = 47.997 amu

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The molar mass of Ammonium Nitrate is 80.043 grams per mole.
3 0
3 years ago
The reaction AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) is a(n) ______________ reaction.
Temka [501]

Answer : The correct option is, (e) precipitation reaction

Explanation :

Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Precipitation reaction : It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.

The insoluble salt that settle down in the solution is known an precipitate.

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It is represented as,

A+BC\rightarrow AC+B

In this reaction, A is more reactive element and B is less reactive element.

Acid-base reaction : It is defined as the reaction in which an acid react with a base to give salt and water as a product.

The given chemical reaction is:

AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)

This reaction is a precipitation reaction in which the two aqueous solution silver nitrate and sodium chloride combined to give sodium nitrate solution and silver chloride as a precipitate. In precipitation reaction there is no changes in oxidation state of the element.

Hence, the correct option is, (e) precipitation

8 0
3 years ago
For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?
harina [27]

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ

→ 5.5 = LnQ → Q= 245 =2.5 * 10^2

6 0
4 years ago
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Greeley [361]

Answer:

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5 0
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