A hydrogen electron is elevated from level 1 to level 2. Another electron is elevated from level 2 to level

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

VARVARA [1.3K]

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

8 0

3 years ago

Given 700 ml of oxygen at 7 ºC and 106.6 kPa pressure, what volume does it take at 27ºC and 66.6 kPa pressure

o-na [289]

Answer:

The volume is 1.2L

Explanation:

Initial volume (V1) = 700mL = 0.7L

Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K

Initial pressure = 106.6kPa = 106600Pa

Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K

Final pressure (P2) = 66.6kPa = 66600Pa

Final volume (V2) = ?

To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.

(P1 × V1) / T1 = (P2 × V2) / T2

solve for V2 by making it the subject of formula,

P1 × V1 × T2 = P2 × V2 × T1

V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)

V2 = 22397193 / 18657990

V2 = 1.2L

The final volume of the gas is 1.2L

6 0

3 years ago

Which radioactive emission has the smallest (least) mass? Question options:

Tems11 [23]

Answer:

The answer is D. gamma rays

Explanation:

A radioactive atom can have three different types of emission:

alpha particles (α) = they have a mass of 4 amu and they have a very low penetrating power.

Beta particles (β) = they have 5x $10^{-4}$ amu and they have an intermediate penetrating power

Gamma rays (γ) = they are not particles basically just energy so its mass is ≈ 0 and its penetrating power is higher

For this reason Gamma emissions (γ) has the smallest mass value.

8 0

3 years ago

A boy walking in the street potential or kenetic energy?

qwelly [4]

Kinetic

because the boy is walking in the street soo it has force to use

7 0

3 years ago

Read 2 more answers

How many formula units of sodium chloride (NaCl) may theoretically be produced from 13.0 g FeCl3?

algol [13]

Answer:

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

Explanation:

Mass of ferric chloride = 13.0 g

Moles of ferric chloride = $\frac{13.0 g}{162.5 g/mol}=0.08 mol$

1 mole of ferric chloride has three moles of chloride ions.Then 0.08 moles of ferric chloride will have :

$3\times 0.08 mol=0.24 mol$ of chloride

$Na^++Cl^-\rightarrow NaCl$

1 mole of sodium ion reacts with 1 mole of chloride ion to form 1 mole of NaCl. Then 0.24 moles of chloride ion will give:

$\frac{1}{1}\times 0.24 mol=0.24 mol$ of NaCl

1 mole = $N_A=6.022\times 10^{23}$ molecules/ atoms

Number of NaCl molecules in 0.24 moles :

$=6.022\times 10^{23}\times 0.24=1.445\times 10^{23} molecules$

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

4 0

3 years ago