Home
Class 10
CHEMISTRY
In the raction Mg + C1 to Mg C1(2)...

In the raction `Mg + C1 to Mg C1_(2)`

A

magnesium is oxidized and `C1_(2)` is reduced

B

magnesium is reduced and `C1_(2)` is oxidized

C

magnesium gains 2 electrons

D

Chlorine loose 1 electron

Text Solution

AI Generated Solution

The correct Answer is:
To analyze the reaction \( \text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2 \), we need to determine the oxidation states of the elements involved and identify which species is oxidized and which is reduced. ### Step-by-Step Solution: 1. **Identify the Reactants and Products**: - Reactants: Magnesium (Mg) and Chlorine gas (\( \text{Cl}_2 \)) - Product: Magnesium chloride (\( \text{MgCl}_2 \)) 2. **Determine the Oxidation States**: - In elemental form, magnesium (\( \text{Mg} \)) has an oxidation state of 0. - In \( \text{MgCl}_2 \), magnesium has an oxidation state of +2. - Chlorine in \( \text{Cl}_2 \) also has an oxidation state of 0. - In \( \text{MgCl}_2 \), each chlorine atom has an oxidation state of -1. 3. **Identify Changes in Oxidation States**: - Magnesium changes from 0 (in \( \text{Mg} \)) to +2 (in \( \text{MgCl}_2 \)), indicating it is oxidized. - Chlorine changes from 0 (in \( \text{Cl}_2 \)) to -1 (in \( \text{MgCl}_2 \)), indicating it is reduced. 4. **Determine Electron Transfer**: - Magnesium loses 2 electrons to go from 0 to +2. - Each chlorine atom gains 1 electron to go from 0 to -1. Since there are 2 chlorine atoms, they collectively gain 2 electrons. 5. **Conclusion**: - Since magnesium is losing electrons, it is the reducing agent (it is oxidized). - Since chlorine is gaining electrons, it is the oxidizing agent (it is reduced). ### Final Answer: - Magnesium (Mg) is oxidized, and chlorine (\( \text{Cl}_2 \)) is reduced in the reaction \( \text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2 \). ---

To analyze the reaction \( \text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2 \), we need to determine the oxidation states of the elements involved and identify which species is oxidized and which is reduced. ### Step-by-Step Solution: 1. **Identify the Reactants and Products**: - Reactants: Magnesium (Mg) and Chlorine gas (\( \text{Cl}_2 \)) - Product: Magnesium chloride (\( \text{MgCl}_2 \)) ...
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL REACTIONS

    MCGROW HILL PUBLICATION|Exercise HIGHER ORDER THINKING QUESTIONS|10 Videos
  • CHEMICAL BONDING

    MCGROW HILL PUBLICATION|Exercise HIGHER ORDER THINKING QUESTIONS|10 Videos
  • MATTER

    MCGROW HILL PUBLICATION|Exercise HIGH ORDER THINKING QUESTIONS|10 Videos

Similar Questions

Explore conceptually related problems

For the given cell, Mg | Mg^(2+) || Cu^(2+) | Cu 1) Mg is cathode 2) Cu is cathode 3) The cell reaction is Mg + Cu^(2+) to Mg^(2+) + Cu 4) Cu is the oxidising agent.

Calculate the DeltaG^(@) and equilibrium constant of the reaction at 27^(@)C Mg+ Cu^(+2) hArr Mg^(+2) + Cu E_(Mg^(2+)//Mg)^(@) = -2.37V, E_(Cu^(2+)//Cu)^(@) = +0.34V

IE_(1) and IE_(2) of Mg are 178 and 348 kcal mol^(-1) . The energy required for the reaction Mg rarr Mg^(2+)+2e^(-) is

Calculate Delta_(r)G^(@) for the reaction : Mg(s)+Cu^(2+)(aq) to Mg^(2+)(aq)+Cu(s) [Given E_(cell)^(@)=+2.71" V ", 1F=96500" C " ]

Consider the cell reaction : Mg(s)+Cu^(2+)(aq) rarr Cu(s) +Mg^(2+)(aq) If E^(c-)._(Mg^(2+)|Mg(s)) and E^(c-)._(Cu^(2+)|Cu(s)) are -2.37 and 0.34V , respectively. E^(c-)._(cell) is

Calculate the Delta G^(@) and equilibrium constant of the rectoi at 27^(@)C Mg+Cu^(+2)- Mg^(+2)+Cu E_(Mg^(2+)//Mg)^(@)=-2.37 V E_(Cu^(2+)//Cu)^(@)=+0.34V

The IE_1 " and " IE_2 of Mg (g) are 740 and 1450 kJ "mol:^(-1) . Calculate the percentage of Mg^(+) (g) and Mg^(2+) (g) if 1g of Mg (g) absorbs 50 kJ of energy.

The solubility of Mg(OH)_(2) is increased by the addition of overset(o+)NH_(4) ion. Calculate a. Kc for the reaction: Mg(OH)_(2) +2overset(o+)NH_(4) hArr 2NH_(3) +2H_(2)O + Mg^(+2) K_(sp) of Mg(OH)_(@) = 6 xx 10^(-12), K_(b) of NH_(3) = 1.8 xx 10^(-5) . b. Find the solubility of Mg(OH)_(2) in a solution containing 0.5M NH_(4)C1 before addition of Mg(OH)_(2) .b

1 mg =