Home
Class 11
CHEMISTRY
Sulphur shows a large number of oxidatio...

Sulphur shows a large number of oxidation states in its compounds such as `H_2S, SO_2, H_2SO_3,Na_2S_2O_3, H_2SO_4, H_2S_2O_7, H_2S_2O_4` and elemental S . In the reactions of sulphur compounds , the oxidation number of sulphur increases and decreases in different reactions and the compounds act as reducing and oxidising agents.
(i) `H_2S +HNO_3 rarr NO+S+H_2O`
(ii) `2Na_2S_2O_3+I_2rarr Na_2S_4O_6+2NaI`
(iii) `H_2SO_4+2HBr rarr 2H_2O+Br_2+SO_2`
(iv) `S_8+12OH^(-) rarr 4S^(2-) +2S_2O_3^(2-) +6H_2O`
Name the oxidising agent reducing agent in reaction (i)

Text Solution

AI Generated Solution

The correct Answer is:
To determine the oxidizing and reducing agents in the reaction \( H_2S + HNO_3 \rightarrow NO + S + H_2O \), we will follow these steps: ### Step 1: Identify the oxidation states of sulfur and nitrogen in the reactants and products. 1. **In \( H_2S \)**: - Hydrogen (H) has an oxidation state of +1. - Let the oxidation state of sulfur (S) be \( x \). - The equation is: \( 2(+1) + x = 0 \) → \( x = -2 \). - Therefore, in \( H_2S \), sulfur is in the -2 oxidation state. 2. **In \( HNO_3 \)**: - Nitrogen (N) has an oxidation state of +5. - Oxygen (O) has an oxidation state of -2. - The equation is: \( x + 3(-2) = 0 \) → \( x = +5 \). - Therefore, in \( HNO_3 \), nitrogen is in the +5 oxidation state. 3. **In the products**: - **For sulfur (S)**: In the elemental form \( S \), the oxidation state is 0. - **For nitrogen (NO)**: Nitrogen has an oxidation state of +2. - The oxidation state of nitrogen in \( NO \) can be calculated as follows: \( x + (-2) = 0 \) → \( x = +2 \). ### Step 2: Determine the changes in oxidation states. - **For sulfur**: - Changes from -2 (in \( H_2S \)) to 0 (elemental S). - This indicates oxidation (loss of electrons). - **For nitrogen**: - Changes from +5 (in \( HNO_3 \)) to +2 (in \( NO \)). - This indicates reduction (gain of electrons). ### Step 3: Identify the oxidizing and reducing agents. - **Oxidizing Agent**: The species that is reduced (gains electrons) and causes oxidation in another species. Here, \( HNO_3 \) is reduced from +5 to +2, thus it is the oxidizing agent. - **Reducing Agent**: The species that is oxidized (loses electrons) and causes reduction in another species. Here, \( H_2S \) is oxidized from -2 to 0, thus it is the reducing agent. ### Conclusion: - **Oxidizing Agent**: \( HNO_3 \) - **Reducing Agent**: \( H_2S \) ---
Promotional Banner

Topper's Solved these Questions

  • REDOX REACTIONS

    MODERN PUBLICATION|Exercise Revision Exercies (True or False Questions )|6 Videos

  • REDOX REACTIONS

    MODERN PUBLICATION|Exercise Revision Exercies (Fill in the Blanks Questions )|8 Videos

  • REDOX REACTIONS

    MODERN PUBLICATION|Exercise NCERT FILE Solved (NCERT - Exemplar Problems) (Long Answer Questions )|6 Videos

  • ORGANIC CHEMISTRY: BASIC PRINCIPLES AND TECHNIQUES

    MODERN PUBLICATION|Exercise (Competition File) MULTIPLE CHOICE QUESTIONS ( based on the given. passage/comprehension )|8 Videos

  • S-BLOCK ELEMENTS ( ALKALI AND ALKALINE EARTH METALS )

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|13 Videos

Similar Questions

Explore conceptually related problems

Sulphur shows a large number of oxidation states in its compounds such as H_2S, SO_2, H_2SO_3,Na_2S_2O_3, H_2SO_4, H_2S_2O_7, H_2S_2O_4 and elemental S . In the reactions of sulphur compounds , the oxidation number of sulphur increases and decreases in different reactions and the compounds act as reducing and oxidising agents. (i) H_2S +HNO_3 rarr NO+S+H_2O (ii) 2Na_2S_2O_3+I_2rarr Na_2S_4O_6+2NaI (iii) H_2SO_4+2HBr rarr 2H_2O+Br_2+SO_2 (iv) S_8+12OH^(-) rarr 4S^(2-) +2S_2O_3^(2-) +6H_2O What type of reaction is reaction (iv) ?

Sulphur shows a large number of oxidation states in its compounds such as H_2S, SO_2, H_2SO_3,Na_2S_2O_3, H_2SO_4, H_2S_2O_7, H_2S_2O_4 and elemental S . In the reactions of sulphur compounds , the oxidation number of sulphur increases and decreases in different reactions and the compounds act as reducing and oxidising agents. (i) H_2S +HNO_3 rarr NO+S+H_2O (ii) 2Na_2S_2O_3+I_2rarr Na_2S_4O_6+2NaI (iii) H_2SO_4+2HBr rarr 2H_2O+Br_2+SO_2 (iv) S_8+12OH^(-) rarr 4S^(2-) +2S_2O_3^(2-) +6H_2O Name the substance which gets (a) reduced and (b) oxidised in reaction (iii)

Sulphur shows a large number of oxidation states in its compounds such as H_2S, SO_2, H_2SO_3,Na_2S_2O_3, H_2SO_4, H_2S_2O_7, H_2S_2O_4 and elemental S . In the reactions of sulphur compounds , the oxidation number of sulphur increases and decreases in different reactions and the compounds act as reducing and oxidising agents. (i) H_2S +HNO_3 rarr NO+S+H_2O (ii) 2Na_2S_2O_3+I_2rarr Na_2S_4O_6+2NaI (iii) H_2SO_4+2HBr rarr 2H_2O+Br_2+SO_2 (iv) S_8+12OH^(-) rarr 4S^(2-) +2S_2O_3^(2-) +6H_2O What is the change in oxidation number of sulphur in reaction (ii)

Sulphur shows a large number of oxidation states in its compounds such as H_2S, SO_2, H_2SO_3,Na_2S_2O_3, H_2SO_4, H_2S_2O_7, H_2S_2O_4 and elemental S . In the reactions of sulphur compounds , the oxidation number of sulphur increases and decreases in different reactions and the compounds act as reducing and oxidising agents. (i) H_2S +HNO_3 rarr NO+S+H_2O (ii) 2Na_2S_2O_3+I_2rarr Na_2S_4O_6+2NaI (iii) H_2SO_4+2HBr rarr 2H_2O+Br_2+SO_2 (iv) S_8+12OH^(-) rarr 4S^(2-) +2S_2O_3^(2-) +6H_2O What is the oxidation number of sulphur in H_2SO_3 and H_2S_2O_4 ?

In the compound Na_(2)S_(2)O_(3) , the oxidation state of sulphur is:

The oxidation number of sulphur in S_8, S_2 F_2, H_2 S and H_2 SO_4 respectively are :

Oxidation states of sulphur in H_2 S_2 O_7 , H_2 SO_3 and H_2 S are respectively