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3 years ago

Do free electrons appear anywhere in the balanced equation for a redox reaction?

2 answers:

7 0

Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.

*Note: Overall redox reactions do not show free electrons because they cancel each other out from both sides of the half-reaction equations.

Andru [333]3 years ago

4 0

Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.

Explanation:

A half reaction is either the chemical reaction or reduction reaction part of an oxidoreduction reaction. A half reaction is obtained by considering the amendment in chemical reaction states of individual substances concerned within the oxidoreduction reaction. Half-reactions are usually used as a way of leveling oxidoreduction reactions.The half-reaction on the anode, wherever chemical reaction happens, is Zn(s) = Zn2+ (aq) + (2e-).

The metal loses 2 electrons to create Zn2+. The half-reaction on the cathode wherever reduction happens is Cu2+ (aq) + 2e- = Cu(s).

Here, the copper ions gain electrons and become solid copper.

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Please help, I really don’t understand this!!!

kap26 [50]

Analysing the Question:

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: for every 1 mole of Glucose, we need 6 moles of Oxygen

Moles of Glucose used in the reaction:

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

Mass of Oxygen required:

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen = 32 / 30

Mass of Oxygen = 1.06 grams

5 0

3 years ago

Microwave radiation has a wavelength on the order of 1.0 cm. Calculate the frequency and the energy of a single photon of this r

denis23 [38]

Answer :

(1) The frequency of photon is, $3\times 10^{10}Hz$

(2) The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

Explanation : Given,

Wavelength of photon = $1.0cm=0.01m$ (1 m = 100 cm)

(1) Now we have to calculate the frequency of photon.

Formula used :

$\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

$\lambda$ = wavelength of photon

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

Now put all the given values in the above formula, we get:

$\nu=\frac{3\times 10^8m/s}{0.01m}$

$\nu=3\times 10^{10}s^{-1}=3\times 10^{10}Hz$ $(1Hz=1s^{-1})$

The frequency of photon is, $3\times 10^{10}Hz$

(2) Now we have to calculate the energy of photon.

Formula used :

$E=h\times \nu$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times (3\times 10^{10}s^{-1})$

$E=1.988\times 10^{-23}J/photon$

The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) Now we have to calculate the energy in J/mol.

$E=1.988\times 10^{-23}J/photon$

$E=(1.988\times 10^{-23}J/photon)\times (6.022\times 10^{23}photon/mol)$

$E=11.97J/mol$

The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

3 0

3 years ago

Read 2 more answers

Classify each one of the following as pure substances or mixtures:

vova2212 [387]

Please refer to the attachment for a complete classification of your specified matter.

Download docx

6 0

3 years ago

Read 2 more answers

Electrons involved in bonding between atoms are _____.

never [62]

Valence électrons = are electrons in the outermost shell that is responsible for the chemical reactions of the atoms

8 0

3 years ago

Plzz someone help me ASAP!

irina1246 [14]

The greater the friction, or rubbing, between particles in any fluid, the higher the viscosity. A fluid with a high viscosity has a large amount of internal friction. As the temperature of a gas increases, friction increases, and so the viscosity of the gas increases. The warmer the gas, the slower it flows.

3 0

3 years ago