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A voltaic cell has one half-cell with a Cu bar in a 1.00 M Cu²⁺ salt, and the other half-cell with a Cd bar in the same volume o

Verizon [17]

The $E^o_{cell}$ of the voltaic cell is 0.743 V. The ΔG°of the given cell is -143.377 kJ and K is 1.35 × 10²⁵

<h3>What is a voltaic cell?</h3>

A voltaic cell often called a galvanic cell, is an electrochemical device that produces electricity through spontaneous redox processes.

It is divided into two distinct half-cells. A half-cell is made up of an electrode (a metal strip, M) dissolved in a solution containing Mn⁺ ions. M can be any metal.

A wire from one electrode to the other connects the two half-cells. Additionally, a salt bridge links the two half-cells.

To solve the question, we need to write the equations of two half-cells

At the anode, oxidation occurs

$Cd(s) \rightarrow Cd^{2+}(aq) + 2e^-$

E° = -0.403 V

At the cathode, reduction occurs

$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

E° = 0.34 V

Overall reaction:

$Cd(s) + Cu^{2+}(aq) \rightarrow Cd^{2+}(aq) + Cu(s)$

We know

$E^o_{cell} = E^o_{cathode} - E^o_{anode}$

= 0.34 -(-0.403) = 0.743 V

Also,

ΔG° = -nF $E^o_{cell}$

Where, n = no of electrons gained or lost

F = Faraday constant

$E^o_{cell}$ = standard potential

ΔG° = -2×96485×0.743 = -143376.71 J = -143.377 kJ

Also,

ΔG° = -RTlnK

-143.377 = -8.314 × 10-3 × 298 × lnK

lnK = 57.87

K = 1.35 × 10²⁵

Hence, The $E^o_{cell}$ of the voltaic cell is 0.743 V. The ΔG°of the given cell is -143.377 kJ and K is 1.35 × 10²⁵

Learn more about Voltaic cell:

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6 0

1 year ago

Sort the phrases based on whether they describe or give an example of diffusion, facilitated diffusion, both, or neither. note:

Bingel [31]

Question:

a. Diffusion

b. Facilitated diffusion

c. Both

d. Neither

1. movement to area of lower concentration

2. movement across a membrane

3. steroid transport into cell

4. requires energy

5. movement assisted by proteins

6. glucose transport into cell

Answer:

The sorting is as follows

a. (1)

b. (5 and 6)

c. (1 and 2)

d. (4)

Explanation:

Diffusion is the movement of particles across a membrane from a high concentration region to one with a lower concentration of the diffusing substance

Here we have the correct sorting as follows

a. Diffusion

3. steroid transport into cell

b. Facilitated diffusion

5. movement assisted by proteins

6. glucose transport into cell

c. Both

1. movement to area of lower concentration

2. movement across a membrane

d. Neither

4. requires energy

8 0

3 years ago

You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2H

KIM [24]

Answer:

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+ pKa1 = 2.14

H2PO4 = HPO4 + H+ pKa2 = 6.86

HPO4 = PO4 + H+ pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

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7 0

3 years ago

Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of

dybincka [34]

Answer: The electronic configuration of the elements are written below.

Explanation:

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

For the given options:

Option a: Carbon (C)

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is $1s^22s^22p^2$

Option b: Phosphorus (P)

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is $1s^22s^22p^63s^23p^3$

Option c: Vanadium (V)

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is $1s^22s^22p^63s^23p^64s^23d^3$

Option d: Antimony (Sb)

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is $1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3$

Option e: Samarium (Sm)

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is $1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6$

Hence, the electronic configuration of the elements are written above.

4 0

3 years ago

An air mattress is filled with 16.5 moles of air. The air inside the mattress has a temperature of 295 K and a gauge pressure of

postnew [5]

PV=nRT
3.5×X=16.5×0.082×295
X= 114 L
The volume of the air mattress is 114 liters.

7 0

3 years ago

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