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ANTONII [103]
2 years ago
14

Two voltaic cells are to be joined so that one will run the other as an electrolytic cell. In the first cell, one half-cell has

Au foil in 1.00 M Au(NO₃)₃, and the other half-cell has a Cr bar in 1.00 M Cr(NO₃)₃. In the second cell, one half-cell has a Co bar in 1.00 M Co(NO₃)₂, and the other half-cell has a Zn bar in 1.00 M Zn(NO₃)₂.(e) If 2.00 g of metal plates out in the voltaic cell, how much metal ion plates out in the electrolytic cell?
Chemistry
1 answer:
Alik [6]2 years ago
7 0

The metal that plates out in the electrolytic cell is zinc. The mass of zinc deposited is 0.996 g.

<h3>What is an electrolytic cell?</h3>

An electrolytic cell converts electrical energy into chemical energy. Unlike galvanic cells, electrons in electrolytic cells move in the opposite direction.

Similar to a galvanic cell, an electrolytic cell has two distinct half-cells: an oxidation half-cell and a reduction half-cell.

In an electrolytic cell, the flow of electrons from the anode, where oxidation takes place, to the cathode, where reduction takes place.

It is fueled by an external source of power (such as a battery) as the reaction in electrolytic cells is not spontaneous, an external source of electrical energy is required.

The reactions for each half cell are given by

Au^{3+}(aq) + 3e^- \rightleftharpoons Au(s) E° = 1.50 V

Cr^{3+}(aq) + 3e^- \rightleftharpoons Cr(s) E° = –0.74 V

Co^{2+}(aq) + 2e^- \rightleftharpoons Co(s) E° = –0.28 V

Zn^{2+}(aq) + 2e^- \rightleftharpoons Zn(s) E° = –0.76 V

According to Faraday's law, when two or more electrolytic cells are connected in series the mass of the substance deposited is directly proportional to its equivalent mass.

Equivalent mass of Au, E_{Au} = Molar mass/no. of moles of electrons transferred

                                      = \frac{197}{3} = 65.67

Equivalent mass of Zinc, E_{Zn} = 65.41/2 = 32.705

\frac{m_{Zn}}{m_{Au }}  = \frac{E_{Zn}}{E_{Au}}

m_{Zn} = \frac{32.705}{65.67} \times 2.00 = 0.996 g

Hence, the mass of Zn deposited is 0.996 g.

Learn more about an electrolytic cell:

brainly.com/question/19854746

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Answer: 9.68 x 10^10 grams.

Explanation:

Given that:

Mass of CO2 = ?

Number of molecules of CO2 = 2.2x10^9 molecules

Molar mass of CO2 = ? (let unknown value be Z)

For the molar mass of CO2: Atomic mass of Carbon = 12; Oxygen = 16

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Apply the formula:

Number of molecules = (Mass of CO2 in grams/Molar mass)

2.2x10^9 molecules = Z/44g/mol

Z = 2.2x10^9 molecules x 44g/mol

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Answer: 8 dots would be needed to represent the outer electrons for an atom of neon in an electron dot diagram.

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Explanation:

From first source, kinetic energy (K.E_{1}) ejected is 1 eV and wavelength of light is \lambda.

From second source, kinetic energy (K.E_{2}) ejected is 4 eV and wavelength of light is \frac{\lambda}{2}.

Relation between work function, wavelength, and kinetic energy is as follows.

                   K.E = \frac{hc}{\lambda} - \phi

where,        h = Plank's constant = 6.63 \times 10^{-34} J.s

                   c = speed of light = 3 \times 10^{8} m/s

Also, it is known that 1 eV = 1.6 \times 10^{-19} J

Therefore, substituting the values in the above formula as follows.

  • From first source,

                      K.E_{1} = \frac{hc}{\lambda} - \phi  

            1 eV = 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi    

    1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi        ........... (1)

  • From second source,

                  K.E_{2} = \frac{hc}{\lambda} - \phi  

          4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi        

                 6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi        ........... (2)        

Now, divide equation (2) by 2. Therefore, it will become

       {6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}

                3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}   ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

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                    \phi = 3.2 \times 10^{-19}

                          = 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

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