Consider the following balanced redox reaction: 2CrO2-(aq) + 2H2O(l) + 6ClO-(aq) LaTeX: \longrightarrow⟶ 2CrO42-(aq) + 3Cl2(g) + 4OH-(aq) 1. Which species is being oxidized? 2. Which species is being reduced? 3. Which species is the oxidizing agent? 4. Which species is the reducing agent? 5. How many electrons are being transferred? Hint: If you were to balance this equation how many electrons would be in each half-reaction? That is how many electrons are transferred.
Answers
Answer:
1. Chromium
2. Chlorine.
3. Chlorine.
4. Chromium.
5. 12 electrons.
Explanation:
Hello,
In this case, the given reaction with the appropriate oxidation states turns out:
In such a way, the oxidation half-reaction is written for chromium as the reducing agent so it is oxidized from +3 to +6, nonetheless, since there are two chromiums undergoing such change, 6 electrons are being transferred as shown below:
On the other hand, chlorine's reduction half-reaction as the oxidizing agent result from the transfer of 6 electrons as well from +1 to 0, nonetheless, there are 6 chlorines undergoing such change:
Therefore, there are 12 electrons that are being transferred, 6 for chromium and 6 for chlorine.
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Final answer:
In the given balance redox reaction, ClO- is the species being oxidized, CrO2- is the species being reduced. CrO2- also serves as the oxidizing agent, and ClO- serves as the reducing agent. A total of 6 electrons are transferred in this reaction.
Explanation:
In the given balanced redox reaction: 2CrO2- (aq) + 2H2O(l) + 6ClO- (aq) ⟶ 2CrO42- (aq) + 3Cl2(g) + 4OH- (aq), the species being oxidized is ClO- since it's gaining oxygen and losing electrons. The species being reduced is CrO2- since it's gaining electrons. The oxidizing agent, which is the substance that causes oxidation by accepting electrons, is CrO2-. The reducing agent, which causes reduction by losing electrons, is ClO-. The number of electrons being transferred in this reaction would be 6, as determined by balancing the half reactions for oxidation and reduction.
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Related Questions
What is the empirical formula of a compound that is 64.3 % c, 7.2 % h, and 28.5 % o by mass?
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Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 C, as the heat provided when 100g of steam is cooled rom 110C to 100C.
Answers
Answer:
(a) Iron is being oxidized.
(d) Sulfur is being reduced.
Explanation:
Let's consider the following redox reaction.
8 Fe(s) + S₈(s) → 8 FeS(s)
Iron is being oxidized according to the following oxidation half-reaction:
Fe(s) → Fe²⁺(s) + 2 e⁻
Sulfur is being reduced according to the following reduction half-reaction:
S₈(s) + 16 e⁻ → 8 S²⁻(s)
Answers
anonymous 5 years ago
First you chlorinate it in presence of light.
C2H6 + Cl2 ---hv -> C2H5Cl + HCl
Then you add aqeuos KOH to get C2H5OH
C2H5Cl+KOH-> C2H5OH+ KCl.
Then you add KMnO4 to get the rquired compound.
C2H5OH ----KMnO4 ---> CH3COOH.
Answers
Answer:
Inorganic impurities present in the bulk of a liquid such as KCl tend to increase the surface tension of water.
Explanation:
As potassium chloride (KCl) dissolves in water, the ions are hydrated. ... When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them.
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Answers
Answer:
a) D = 33.44 Lbmol/h
⇒ B = 62.56 Lbmol/h
b) D = 16.848 Kmol/h
⇒ B = 28.152 Kmol/h
Explanation:
global balance:
- F = D + B........................(1)
∴ F = 100 Lbmol/h
balance per component:
A: 0.4*F = 0.9*D + 0.1*B = 0.4*100 = 40 Lbmol/h..............(2)
B: 0.6*F = 0.1*D + 0.9*B = 0.6*100 = 60 Lbmol/h..............(3)
from (2):
⇒ 0.9*D = 40 - 0.1*B
⇒ D = ( 40 - 0.1*B ) / 0.9............(4)
(4) in (3):
⇒ 0.1*((40-0.1*B)/0.9) + 0.9*B = 60
⇒ B = 62.56 Lbmol/h............(5)
(5) in (1):
⇒ D = 100 - B
⇒ D = 37.44 Lbmol/h
∴ Lbmol = 0.45 Kmol
⇒ B = 62.56 Lbmol/h * ( 0.45 Kmol/ Lbmol ) = 28.152 Kmol/h
⇒ D = 37.44 Lbmol/h * ( 0.45 Kmol/h ) = 16.848 Kmol/h
Answers
Answer:
To determine the value of Kp for the given equilibrium, we need to use the partial pressures of the gases involved.
In the balanced equation: 2 HI (g) ⇌ H₂ (g) + I₂ (g), the stoichiometric coefficients are 2, 1, and 1 respectively.
At equilibrium, the expression for Kp is given by:
Kp = (P(H₂) * P(I₂)) / (P(HI)²)
Using the provided partial pressures:
P(HI) = 1.9 atm
P(H₂) = 7.9 atm
P(I₂) = 2.3 atm
Substituting these values into the expression for Kp:
Kp = (7.9 * 2.3) / (1.9²)
Kp ≈ 19.5 / 3.61
Calculating the result:
Kp ≈ 5.4
Therefore, the value of Kp for the given equilibrium is approximately 5.4.
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[Ar]3d^5
Check all that apply.
A- Fe2+
B- Fe3+
C- Mn2+
D- V+
E- Sc2+
Answers
Answer:
Explanation:
Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom. The electrons are filled according to Afbau's rule in order of increasing energies.
The electronic configuration for given elements is as follows:
Final answer:
The ions Fe2+ and Mn2+ have the ground-state electron configuration [Ar]3d^5.
Explanation:
The ground-state electron configuration [Ar]3d^5 indicates a level of electrons in 3d subshell after the Argon core electron configuration. Now, iron (Fe) has a base atomic configuration of [Ar]3d^6 4s2. When it loses 2 electrons (to form Fe2+), it tends to lose from both the 3d and the 4s sublevels, giving [Ar]3d^5 (which is our required configuration).
However, it's also important to consider Manganese (Mn), which has a base configuration of [Ar]3d^5 4s2. It usually loses 2 electrons from the 4s sublevel first when it forms Mn2+ which results in a configuration [Ar]3d^5.
So, the two ions with the electron configuration [Ar]3d^5 are Fe2+ and Mn2+.
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