Solved Examples: Redox Reactions & Electrochemistry Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE

JEE: Solved Examples: Redox Reactions & Electrochemistry Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE

The document Solved Examples: Redox Reactions & Electrochemistry Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2022.
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Question 1: Given
Solved Examples: Redox Reactions & Electrochemistry Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE

Based on the data given above, strongest oxidising agent will be.    (IIT JEE-2013)
(a) Cr3+ 
(b) Mn2+
(c) MnO4-
(d) Cl- 
Answer: c
Solution:
More positive value of standard reduction potential indicated the poor tendency of the species to undergo reduction. In the given data, MnO4- has the highest positive value of standard reduction potential this means that it has least tendency to undergo reduction and thus highest tendency to undergo oxidation among the given species, thus it is the strongest oxidising agent.
Hence, the correct option is c.

Question 2: The reaction of white phosphorous with aqueous NaOH gives phosphine along with another phosphorous containing compound. The reaction type; the oxidation on states of phosphorus in phosphine and the other product are respectively    (IIT JEE -2012)
(a) redox reaction; −3 and −5
(b) redox reaction; +3 and +5
(c) disproportionation reaction; −3 and +5
(d) disproportionation reaction; −3 and +3
Answer: c
Solution: The Balanced Reaction:
Solved Examples: Redox Reactions & Electrochemistry Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE
It is clear from the balanced reaction that it is a disproportionation reaction as P undergo both oxidation as well as reduction in this reaction.
NaH2PO2 formed in this reaction is less stable and decompose to from two products
Na2HPO4 and PH3. Oxidation number of P in Na2HPO4 is +5.
Hence, the correct option is C.

Question 3: The solubility product (Ksp; moldm–9) of MX2 at 298 K based on the information available for the given concentration cell is (take 2.303 ×R ×298/F = 0.059 V).    (IIT JEE -2012)
(a) 1 ×10–15
(b) 4 ×10–15
(c) 1 ×10–12
(d) 4 ×10–12
Answer: c
Solution:
Solved Examples: Redox Reactions & Electrochemistry Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE
Hence, the correct option is C.

The document Solved Examples: Redox Reactions & Electrochemistry Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2022.
All you need of JEE at this link: JEE

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