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Solution: Identify the oxidizing and reducing agents in the
Chapter 4, Problem 59P(choose chapter or problem)
Identify the oxidizing and reducing agents in the following:
\(\mathrm{BH}+(\mathrm{aq})+\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2}(\mathrm{aq})+3 \mathrm{SO}_{3}{ }^{2}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{SO}_{4}{ }^{2-}(\mathrm{aq})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
\(\mathrm{NO}_{3}{ }^{-}(\mathrm{aq})+4 \mathrm{Zn}(\mathrm{s})+7 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 4 \mathrm{Zn}(\mathrm{OH})_{4}{ }^{2}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq})\)
Equation Transcription:
8H+(aq) + Cr2O72-(aq) + 3SO32-(aq) 2Cr3+(aq) + 3SO42-(aq) + 4H2O(l)
NO3-(aq) + 4Zn(s) + 7OH-(aq) 4Zn(OH)42-(aq) + NH3(aq)
Text Transcription:
8H+(aq) + Cr_2O_7^2-(aq) + 3SO_3^2-(aq) rightarrow 2Cr^3+(aq) + 3SO_4^2-(aq) + 4H_2O(l)
NO_3-(aq) + 4Zn(s) + 7OH-(aq) rightarrow 4Zn(OH)_4^2-(aq) + NH_3(aq)
Questions & Answers
QUESTION:
Identify the oxidizing and reducing agents in the following:
\(\mathrm{BH}+(\mathrm{aq})+\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2}(\mathrm{aq})+3 \mathrm{SO}_{3}{ }^{2}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{SO}_{4}{ }^{2-}(\mathrm{aq})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
\(\mathrm{NO}_{3}{ }^{-}(\mathrm{aq})+4 \mathrm{Zn}(\mathrm{s})+7 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 4 \mathrm{Zn}(\mathrm{OH})_{4}{ }^{2}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq})\)
Equation Transcription:
8H+(aq) + Cr2O72-(aq) + 3SO32-(aq) 2Cr3+(aq) + 3SO42-(aq) + 4H2O(l)
NO3-(aq) + 4Zn(s) + 7OH-(aq) 4Zn(OH)42-(aq) + NH3(aq)
Text Transcription:
8H+(aq) + Cr_2O_7^2-(aq) + 3SO_3^2-(aq) rightarrow 2Cr^3+(aq) + 3SO_4^2-(aq) + 4H_2O(l)
NO_3-(aq) + 4Zn(s) + 7OH-(aq) rightarrow 4Zn(OH)_4^2-(aq) + NH_3(aq)
ANSWER:
Solution 59P
Here we have to identify the oxidizing and reducing agents in the following:
Reducing agent:- Loss of electron is known as oxidation reaction and the atom or compound which losses electron is called as reducing agent.
Oxidizing agent:- Gain of electron is known as reduction reaction and the atom or compound which gain electron is called as oxidizing agent.
Step 1
(a) 8H+ (aq) + Cr2O72- (aq) + 3 SO32- (aq) → 2 Cr3+ (aq) + 3 SO42- (aq) + 4 H2O (l)
In order to find out the oxidizing and reducing agent 1st we have to write the oxidation state of each atoms. Then we will come to know from the reactant side to product which atom has gain electron or which one losses.
8H+ (aq) + Cr2O72- (aq) + 3 SO32- (aq) → 2 Cr3+ (aq) + 3 SO42- (aq) + 4 H2O (l)
H = +1 Cr = +6 S=+4 Cr = +3 S =+6 H = +1
O = -2 O = -2 O = -2 O = -2
While combining the reactant and the product side, it has been found that the oxidation number of H and O remains same.
The oxidation state of Cr changes from +6 to +3 (reduction reaction) hence Cr2O72- is the oxidizing agent. The oxidation state of S changes from +4 to +6 (oxidation reaction) hence S is the reducing agent.