Convert the following balanced half-reaction in acid to a balanced half-reaction
ID: 1055684 • Letter: C
Question
Convert the following balanced half-reaction in acid to a balanced half-reaction in base: C_r_2O_7^2-(aq) + 14H^+ + 6e^-(aq) rightarrow 2Cr^3+(aq) + 7H_2O(l) How many OH^- are added to each side? After adding the OH^- and simplifying how many water molecules are on the left? Using smallest integer coefficients balance the half-reaction given below in acid. How many H^+ are on the right? As_2S_3(s) rightarrow AsO_4^3-(aq) + SO_4^2-(aq) How many electrons are added to the right? Convert the balanced half-reaction to one balanced in base. How many OH^- are added to each side? After simplifying, how many water molecule remain on the right side of the equation?Explanation / Answer
Cr2O72-(aq) + 14H+ + 6e-(aq) ---> 2Cr3+ (aq) + 7H2O(l)
Add 14OH- to each side.
Cr2O72-(aq) + 14H+ + 14OH- + 6e-(aq) ---> 2Cr3+ (aq) + 7H2O(l) + 14OH-
or, Cr2O72-(aq) + 14H2O(l) + 6e-(aq) ---> 2Cr3+ (aq) + 7H2O(l) + 14OH- ..[ as 14H+ + 14OH- = 14H2O]
or, Cr2O72-(aq) + 7H2O(l) + 6e-(aq) ---> 2Cr3+ (aq) + 14OH-
So, answer of 16: 14OH- are added to each side.
Answer of 17: 7 water molecules are on the left.
As2S3(s) ---> AsO43-(aq) + SO42-(aq)
Step 1: Add 2 before AsO43- on the right.
As2S3(s) ---> 2AsO43-(aq) + SO42-(aq)
Step 2: Add 3 before SO42- on the right.
As2S3(s) ---> 2AsO43-(aq) + 3SO42-(aq)
Step 3: Add 28e(As goes from +3 to +5, for 2As 4 electrons will be released. S goes from -2 to +6, for 3S total 24 electrons will be released) to the right to balance the electrons.
As2S3(s) ---> 2AsO43-(aq) + 3SO42-(aq) + 28e
Step 4: Total O atom on the right is 20. So add 20 H2O on the left.
As2S3(s) + 20 H2O(l) ---> 2AsO43-(aq) + 3SO42-(aq) + 28e
Step 5: Now add 40H+ on the right.
As2S3(s) + 20 H2O(l) ---> 2AsO43-(aq) + 3SO42-(aq) + 40H+ + 28e
Now the equation is balanced.
Answer of question 18: 40 H+ are on the right.
Answer of question 19: 28 electrons
Answer of question 20:
As2S3(s) + 20 H2O(l) ---> 2AsO43-(aq) + 3SO42-(aq) + 40H+ + 28e
As2S3(s) + 20 H2O(l) + 40 OH- ---> 2AsO43-(aq) + 3SO42-(aq) + 40H+ + 40 OH- + 28e
or, As2S3(s) + 20 H2O(l) + 40 OH- ---> 2AsO43-(aq) + 3SO42-(aq) + 40 H2O + 28e
or, As2S3(s) + 40 OH- ---> 2AsO43-(aq) + 3SO42-(aq) + 20 H2O + 28e
40 OH- are added to each side.
Answer of 21: 20 H2O remains.