A reduced element (which gains electrons) and an oxidized element are required for redox reactions (gives electrons). It is not a redox reaction if we lack both of them (an element can not receive electrons if no element gives electrons and vice versa).
A reduced half and an oxidized half, which always occur together, make up redox processes. While the oxidized half experiences electron loss and an increase in oxidation number, the reduced half obtains electrons and the oxidation number declines. The mnemonic devices OIL RIG, which stand for "oxidation is loss" and "reduction is gain," are simple ways to memorize this. In a redox process, the total number of electrons stays constant. In the reduction half reaction, another species absorbs those that were released in the oxidation half reaction.
In a redox reaction, two species exchange electrons, and they are given unique names:
- The ion or molecule that accepts electrons is called the oxidizing agent - by accepting electrons it oxidizes other species.
- The ion or molecule that donates electrons is called the reducing agent - by giving electrons it reduces the other species.
Hence, what is oxidized is the reducing agent and what is reduced is the oxidizing agent.
<h3>
What is the purpose of oxidizing agents and reducing agents?</h3>
By reducing other compounds and shedding electrons, a reducing agent raises its oxidation state. An oxidizing agent gets electrons by oxidizing other compounds; as a result, its oxidation state lowers.
<h3>
What is a redox reaction?</h3>
Oxidation-reduction (or "redox") reactions are chemical processes in which electrons are exchanged between two substances. An oxidation-reduction reaction is any chemical process in which a molecule, atom, or ion alters the number of electrons it has, hence increasing or decreasing its oxidation state.
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Answer:
Eat food
Explanation:
Food contain protein, carbohydrates, fats etc. These nutrients undergoes metabolic process and produced energy in cellular respiration.
There are two types of respiration:
1. Aerobic respiration
2. Anaerobic respiration
Aerobic respiration
It is the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.
Glucose + oxygen → carbon dioxide + water + 38ATP
Anaerobic Respiration
It is the breakdown of glucose molecule in the absence of oxygen and produce small amount of energy. Alcohol or lactic acid and carbon dioxide are also produced as byproducts.
Glucose→ lactic acid/alcohol + 2ATP + carbon dioxide
This process use respiratory electron transport chain as electron acceptor instead of oxygen. It is mostly occur in prokaryotes. Its main advantage is that it produce energy (ATP) very quickly as compared to aerobic respiration.
Answer:
Explanation:
I gave this formula as
What you should get - what you got
--------------------------------------------------- * 100
What you should get
Suppose you are talking about the specific heat of copper and you did a lab and did it fairly well and the specific heat came out to be 0.33
That's what you got, 0.33
What you should get is 0.35 according to the table you've given us.
% error = 0.35 - 0.33
-------------- * 100
0.35
% error = 0.02 * 100/ 0.35
% error = 2/0.35
% error = 5.71
Now here is the really hard part. You have to decide what a positive error is and what a negative error.
A positive error using this formula means you are too low.
A Negative error means that you are too high, just the opposite of what you might think.
Answer:
(a) Oxidation: Fe(II)(cyt b) ⇌ Fe(III)(cyt b) + e⁻
Reduction: Fe(III)(cyt c1) + e⁻ ⇌ Fe(II)(cyt c1)
(b) K = 750; spontaneous
Explanation:
(a) Half-reactions
Oxidation: Fe(II)(cyt b) ⇌ Fe(III)(cyt b) + e⁻
Reduction: Fe(III)(cyt c1) + e⁻ ⇌ Fe(II)(cyt c1)
Overall: Fe(II)(cyt b) + Fe(III)(cyt c1) ⇌ Fe(III)(cyt b) + Fe(II)(cyt c1)
(b) Equilibrium constant
(i) Standard cell potential
The standard reduction potentials are
E°/V
Fe(III)(cyt c1) + e⁻ ⇌ Fe(II)(cyt c1); 0.25
Fe(III)(cyt b) + e- ⇌ Fe(II)(cyt b); 0.08
For the reaction in Part (a),
<u>E°/V</u>
Fe(II)(cyt b) ⇌ Fe(III)(cyt b) -0.08
Fe(III)(cyt c1) ⇌ Fe(II)(cyt c1) 0.25
Fe(II)(cyt b) + Fe(III)(cyt c1) ⇌ Fe(III)(cyt b) + Fe(II)(cyt c1) 0.17
The standard cell potential is 0.17 V.
(ii) Equilibrium constant
E° and K are positive, so the reaction is spontaneous.
The question only asks regarding the direction of the equilibrium reaction. The general expression of Kp is:
Kp = [PCl₅]/[PCl₃][Cl₂]
The higher the value of K (greater than 1), the more spontaneous the reaction (favors the product side). Otherwise, it favors the reactant side. Since Kp = 0.087 which is less than 1, the direction favors the forward reaction towards the product side.