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spin [16.1K]
3 years ago
5

At which electrode does oxidation occur in a

Chemistry
2 answers:
Verdich [7]3 years ago
7 0

Answer : The correct option is, (3) the anode in both a voltaic cell and an  electrolytic cell.

Explanation :

As we know that there are two types of cell which are, electrolytic cell and voltaic cell.

Electrolytic cell : It is a cell in which the chemical reaction occurs by passing of the current from external source.

In this electrolytic cell, the oxidation occurs at anode which is a positive electrode and reduction occurs at cathode which is a negative electrode.

Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

Hence, the oxidation occur in a  voltaic cell and in an electrolytic cell at the anode in both a voltaic cell and an  electrolytic cell.

Trava [24]3 years ago
5 0
The right answer for the question that is being asked and shown above is that: "(2) the cathode in a voltaic cell and the anode in an electrolytic cell." At the status of electrode does oxidation occur in a voltaic cell and in an electrolytic cell is that the cathode in a voltaic cell and the anode in <span>an electrolytic cell</span>
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The number of protons in every atom of an element
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Answer:

The *atomic* number

Explanation:

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2 years ago
Calculate q (in units of joules) when 1.850 g of water is heated from 22 °C to 33 °C. Report only the numerical portion of your
Minchanka [31]
When q is the heat energy in joules (J)

so, according to this formula, we can get q (in joule unit):

q = M*C*ΔT

when M is the mass of the water sample = 1.85 g

C is the specific heat capacity of water = 4.18 J/g.°C

and Δ T is the difference in temperature (Tf-Ti) = 33 - 22 = 11°C

So, by substitution, we will get the value of q ( in Joule):

∴ q = 1.85 g * 4.18 J/g.°C * 11 °C

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5 0
3 years ago
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3 years ago
A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N
vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

                 2NH₃(g)   ⇄     N₂(g)     +     3H₂(g)

Initially       0.0733

React         0.0733α          α/2                3/2α

Eq     0.0733 - 0.0733α    α/2                0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

3 0
3 years ago
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