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3 years ago

Look at the diagram of an electrolytic cell below.

Chemistry

1 answer:

Ipatiy [6.2K]3 years ago

8 0

<em>The cathode is Y</em>

<h3><em>Further explanation</em></h3>

Electrolysis uses electrical energy to carry out redox reactions that are not spontaneous.

The ions in the solution flowing electrically will move towards to opposite charge of the electrode

The electrolysis material is an electrolyte which can be a solution or a melt.

In positive pole electrolysis cells - the anode is the site of the oxidation reaction, while the negative pole - the cathode is the reduction reaction site.

The result of the reaction in the anode is based on a substance that easily oxidized while the reaction in the cathode is based on a substance that easily reduced.

Electrons (electricity) enter an electrolysis cell through the negative pole (cathode)

The negative ion from the solution will move towards the positive electrode and release the electrons around the positive electrode (oxidation) and the electrons flow to the negative pole

Whereas around the negative electrode, there is electron binding and a reduction reaction occurs

So if we see the picture the cathode is Y

<h3><em>Learn more</em></h3><h3><em>reaction related to electrochemistry brainly.com/question/3461108</em></h3>

<h3><em>Answer details </em></h3>

Grade: Senior High School

Subject: Chemistry

Chapter: Electrochemistry

Keywords: cathode, anode, oxidation, reduction, negative pole, electrode

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Read 2 more answers

Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:

Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

$\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )$ ,

where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from $n_1$ to $n_2$ and $R_h$ = 109677 is the Rydberg Constant. Here $n_1$ and $n_2$ represents the transitions.

(a) $n_1$ =2 to $n_2$ = infinity

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )$ = 109677/4 [since 1/infinity = 0] Therefore, $\lambda$ = 4 / 109677 = 0.00003647 m

(b) $n_1$ =4 to $n_2$ = 20

$\frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2} )$ = 6580.62

Therefore, $\lambda$ = 1 / 6580.62 = 0.000152 m

(c) $n_1$ =3 to $n_2$ = 10

$\frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2} )$ = 11089.56

Therefore, $\lambda$ = 1 / 11089.56 = 0.00009 m

(d) $n_1$ =2 to $n_2$ = 1

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2} )$ = - 82257.75

Therefore, $\lambda$ = 1 /82257.75 = - 0.0000121 m

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

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