Hello
Good evening sir/mam
PLZZ help me by answering these two questions
It's a request
PLZZ answer although these are not from any chapter
Dear student,
It is our pleasure and duty to help you , but it would not be possible if you are not specific about your request. You have posted seven questions but you havenot specified which ones to answer!! Next time, please ask a question at a time, or please specify which ones to be answered
Here, i see that you havenot marked an answer for question 4 , about the reducing reaction of hydrogen sulphide. I assume that this is the question you wish me to answer. So i am proceeding with the answer to that specific question.
Hydrogen sulphide is a good reducing agent. A substance is said to be reducing if it donates hydrogen or electrons and reduces the other compound resulting decrease in oxidation number.
In reaction 1, Cd(NO3)2 , Cd has oxidation number of +2. The product CdS also has Cd at +2 oxidation state. Hence hydrogen sulphide has not reduced any component.
The case is same in 2nd reaction where Cu is not reduced as it has same oxidation number +2 in both CuSO4 and in CuS.
Reaction 4 also shows similar outcomes, since Pb has oxidation state of +2 in PbSO4 aand PbS.
But in reaction 3, Fe has oxidation number of +3 in FeCl3 , but it reduces to +2 in FeCl2 , tthus showing the reducing character of hydrogen sulphide.
Hence , answer is 3.
It is our pleasure and duty to help you , but it would not be possible if you are not specific about your request. You have posted seven questions but you havenot specified which ones to answer!! Next time, please ask a question at a time, or please specify which ones to be answered
Here, i see that you havenot marked an answer for question 4 , about the reducing reaction of hydrogen sulphide. I assume that this is the question you wish me to answer. So i am proceeding with the answer to that specific question.
Hydrogen sulphide is a good reducing agent. A substance is said to be reducing if it donates hydrogen or electrons and reduces the other compound resulting decrease in oxidation number.
In reaction 1, Cd(NO3)2 , Cd has oxidation number of +2. The product CdS also has Cd at +2 oxidation state. Hence hydrogen sulphide has not reduced any component.
The case is same in 2nd reaction where Cu is not reduced as it has same oxidation number +2 in both CuSO4 and in CuS.
Reaction 4 also shows similar outcomes, since Pb has oxidation state of +2 in PbSO4 aand PbS.
But in reaction 3, Fe has oxidation number of +3 in FeCl3 , but it reduces to +2 in FeCl2 , tthus showing the reducing character of hydrogen sulphide.
Hence , answer is 3.