All categories

1 year ago

Consider the following voltaic cell:(d) At which electrode are electrons consumed?

Chemistry

1 answer:

11Alexandr11 [23.1K]1 year ago

4 0

Electrons are consumed at the cathode.

Cathode

the cathode is referred to as the electrode where reduction takes place. In an electrochemical cell, this is typical. Here, the cathode is negative because the cell's electrical energy supply causes chemical molecules to break down. It can, however, also be advantageous, as in a galvanic cell, where a chemical reaction results in the production of electrical energy.

A cathode is often referred to as a hot cathode or a cold cathode. A hot cathode is one that has been heated by a filament in order to emit electrons by thermionic emission, as opposed to a cold cathode, which has not been heated by a filament. If a cathode emits more electrons than those produced by thermionic emission alone, it is typically labeled as "cold."

Learn more about cathode here:-

brainly.com/question/19261112

#SPJ4

You might be interested in

Light strikes a beautiful white perch under water. This light is reflected back to your eyes allowing you to see the fish. As th

icang [17]

The answer is: <span>Light strikes a beautiful white perch under water. This light is reflected back to your eyes allowing you to see the fish. As the light left the water it changes speed causing the light to bend away from the normal.
Hope this helps!
(got answer from </span>https://quizlet.com/5474123/chapter-13-light-flash-cards/ if you need more help)

7 0

2 years ago

What is the best way to package and preserve a questioned document?

Zanzabum

In a plastic sheet protector

5 0

3 years ago

Read 2 more answers

Consider the reaction 2Al(OH)3(s)→Al2O3(s)+3H2O(l) with enthalpy of reaction ΔHrxn∘=21.0kJ/mol What is the enthalpy of formation

bekas [8.4K]

Google knows your answer

8 0

2 years ago

Read 2 more answers

You drop a rock weighing 23.2 g into a graduated cylinder that contains 55 mL. The level of the water rises to 62 mL. What is th

ra1l [238]

The density of the rock is 3.314g/mL

CALCULATE DENSITY:

According to this question, a rock weighs 23.2g. After dropping the rock into a graduated cylinder containing 55mL of water, the level changes to 62mL.

This means that the volume of the rock can be calculated as follows:

Volume of rock = 62mL - 55mL

Volume of rock = 7mL

Density can be calculated using the formula as follows:

Density = mass ÷ volume

Density = 23.2 ÷ 7

Density = 3.314g/mL

Therefore, the density of the rock is 3.314g/mL

Learn more: brainly.com/question/6034174?referrer=searchResults

4 0

2 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

Learn more:

6 0

2 years ago