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Question: Which of the following reactions is a redox reaction? (a) K2CrO4 + BaCl2 ? BaCrO4 + 2KCl (b) Pb22+ + 2Br- ? PbBr (c) Cu + S ? CuSAnswer (a) only (b) only (c) only (b) and (c) (a) and (c)

Which of the following reactions is a redox reaction? (a) K2CrO4 + BaCl2 ? BaCrO4 + 2KCl (b) Pb22+ + 2Br- ? PbBr (c) Cu + S ? CuSAnswer (a) only (b) only (c) only (b) and (c) (a) and (c)

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General guidance

Concepts and reason

The concept used to solve the problem is based on a redox reaction.

A redox reaction is a reaction in which oxidation and reduction reactions occur simultaneously. Fundamentals

Oxidation is a process that involves the addition of oxygen or any other electronegative element, or the removal of an electropositive element or hydrogen. It is the increase in the number of atoms.

Reduction is a process that involves the addition of hydrogen or any electropositive element or the removal of oxygen or any other electronegative element. Show more First Step | All Steps | Answer Only

Step-by-step

Step 1 of 2

Reaction (a) is given as follows:

K2CrO4+BaCl2→BaCrO4+2KCl{{\rm{K}}_{\rm{2}}}{\rm{Cr}}{{\rm{O}}_{\rm{4}}} + {\rm{BaC}}{{\rm{l}}_2} \to {\rm{BaCr}}{{\rm{O}}_{\rm{4}}} + 2{\rm{KCl}}K2​CrO4​+BaCl2​→BaCrO4​+2KCl

Write the oxidation states of each species in the above reaction as follows:

K2+1Cr+6O4−2+Ba+2Cl2−1→Ba+2Cr+6O4−2+2K+1Cl−1{\mathop {\rm{K}}\limits^{ + 1} _{\rm{2}}}\mathop {{\rm{Cr}}}\limits^{ + 6} {\mathop {\rm{O}}\limits^{ – 2} _{\rm{4}}} + \mathop {{\rm{Ba}}}\limits^{ + 2} {\mathop {{\rm{Cl}}}\limits^{ – 1} _2} \to \mathop {{\rm{Ba}}}\limits^{ + 2} \mathop {{\rm{Cr}}}\limits^{ + 6} {\mathop {\rm{O}}\limits^{ – 2} _{\rm{4}}} + 2\mathop {\rm{K}}\limits^{ + 1} \mathop {{\rm{Cl}}}\limits^{ – 1} 2K+1​Cr+64O−2​+Ba+22Cl−1​→Ba+2Cr+64O−2​+2K+1Cl−1

In the above reaction, the oxidation number of barium, chlorine, potassium, chromium, and oxygen does not change after the formation of products. Thus, it is not a redox reaction.

Reaction (b) is given as follows:

Pb22++2Br−→2PbBr{\rm{Pb}}_{\rm{2}}^{{\rm{2 + }}} + 2{\rm{B}}{{\rm{r}}^ – } \to 2{\rm{PbBr}}Pb22+​+2Br−→2PbBr

Write the oxidation states of each species in the above reaction as follows:

Pb22++1+2Br−1→2Pb+1Br−1\mathop {{\rm{Pb}}_2^{2 + }}\limits^{ + 1} + 2\mathop {{\rm{Br}}}\limits^{ – 1} \to 2\mathop {{\rm{Pb}}}\limits^{ + 1} \mathop {{\rm{Br}}}\limits^{ – 1} Pb22+​+1​+2Br−1→2Pb+1Br−1

In the above reaction, the oxidation number of lead and bromine does not change after the formation of products. Thus, it is not a redox reaction.Explanation | Common mistakes | Hint for next step

In reaction (a), the oxidation numbers of barium, chlorine, potassium, chromium, and oxygen have neither decreased nor increased. Therefore, the reaction cannot be regarded as a redox reaction. In reaction (b), the oxidation number of lead and bromine does not change. Thus, it is not a redox reaction.

Be careful while analyzing the change in oxidation as it can lead to confusion and can result in wrong interpretation.

Determine the redox reaction.

Step 2 of 2

Reaction (c) is given as follows:

Cu+S→CuS{\rm{Cu}} + {\rm{S}} \to {\rm{CuS}}Cu+S→CuS

Write the oxidation states of each species in the above reaction as follows:

Cu0+S0→Cu+2S−2\mathop {{\rm{Cu}}}\limits^0 + \mathop {\rm{S}}\limits^0 \to \mathop {{\rm{Cu}}}\limits^{ + 2} \mathop {\rm{S}}\limits^{ – 2}Cu0+S0→Cu+2S−2

In the above reaction, the oxidation number of copper is increasing and sulfide is decreasing. Thus, it is a redox reaction.

The redox reaction is written as follows:

Cu+S→CuS{\rm{Cu}} + {\rm{S}} \to {\rm{CuS}}Cu+S→CuS
Explanation | Common mistakes

Reaction (c) is a redox reaction as the oxidation number of copper increases from 000 to +2 + 2+2 and the oxidation number of sulfide decreases from 000 to −2 – 2−2 . Thus, it is a redox reaction.

Do not consider the oxidation number copper in CuS{\rm{CuS}}CuS as 000 .

Answer

The redox reaction is written as follows:

Cu+S→CuS{\rm{Cu}} + {\rm{S}} \to {\rm{CuS}}Cu+S→CuS

Answer only

The redox reaction is written as follows:

Cu+S→CuS{\rm{Cu}} + {\rm{S}} \to {\rm{CuS}}Cu+S→CuS



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