There are several arsenic minerals, two of which are claudetite (As2O3) and orpiment (As2S3). Wrtie a balanced half cell reaction between these two minerals, in an aqueous solution.
The half cell reaction is: As2O3 + 3 SO4^2- + 30H+ +24e- <--> As2S3 + 15H2O
How do you know that As2O3 will be on the left side of the equation than the right?
How do you know to add SO4^2- to the equation?
How do you know to add the 24e- to the left side of the equation?
If someone could show me steps to explain these questions, it\'d help me out a lot. Thank you.
Solution
The half cell reaction is: As 2 O 3 + 3 SO 4 2- + 30H + +24e - <--> As 2 S 3 + 15H 2 O
How do you know that As 2 O 3 will be on the left side of the equation than the right?
Ans:- As 2 O 3 is soluble in dilute acids and this reaction takes place in acidic condition that is why As 2 O 3 is on left hand side. As 2 S 3 is not soluble in acid at room temperature that is why this will be on right hand side of the equation.
How do you know to add SO 4 2- to the equation?
Ans:- As this reaction takes place in acidic medium ,in the presence of sulphuric acid (H 2 SO 4 ) that is why ionised form of sulphuric acid i.e.SO 4 2- is added to the solution.
How do you know to add the 24e- to the left side of the equation?
Ans:- To balance the charge of the half cell reaction.There is no charge on the right hand side and a (24H + ) + 3(-2) = (30+) - 6 = 24+ charge on the left side of the equation.so to balance this charge add 24 e - to the equation.Now equation become neutral.
.
There are several arsenic minerals- two of which are claudetite (As2O3.docx
1. There are several arsenic minerals, two of which are claudetite (As2O3) and orpiment (As2S3).
Wrtie a balanced half cell reaction between these two minerals, in an aqueous solution.
The half cell reaction is: As2O3 + 3 SO4^2- + 30H+ +24e- <--> As2S3 + 15H2O
How do you know that As2O3 will be on the left side of the equation than the right?
How do you know to add SO4^2- to the equation?
How do you know to add the 24e- to the left side of the equation?
If someone could show me steps to explain these questions, it'd help me out a lot. Thank you.
Solution
The half cell reaction is: As 2 O 3 + 3 SO 4
2- + 30H + +24e - <--> As 2 S 3 + 15H 2 O
How do you know that As 2 O 3 will be on the left side of the equation than the right?
Ans:- As 2 O 3 is soluble in dilute acids and this reaction takes place in acidic condition that is
why As 2 O 3 is on left hand side. As 2 S 3 is not soluble in acid at room temperature that is why
this will be on right hand side of the equation.
How do you know to add SO 4
2-
to the equation?
Ans:- As this reaction takes place in acidic medium ,in the presence of sulphuric acid (H 2 SO 4 )
that is why ionised form of sulphuric acid i.e.SO 4
2-
is added to the solution.
How do you know to add the 24e- to the left side of the equation?
Ans:- To balance the charge of the half cell reaction.There is no charge on the right hand side
and a (24H +
) + 3(-2) = (30+) - 6 = 24+ charge on the left side of the equation.so to balance this
charge add 24 e -
to the equation.Now equation become neutral.
