Redox stoichiometry

  • Thread starter Suy
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  • #1
Suy
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Homework Statement



a 0.245 mol/L silver chloride solution is mixed with 50.0 mL of 0.200 mol/L zinc iodide solution. what volume of silver chloride solution is needed for a complete reaction?
(answer is 81.6mL)

Homework Equations





The Attempt at a Solution


Ag+ is the strongest OA,
I- is the strongest RA
2Ag+2e-->2Ag(s)
2I- -->2e+I2
2Ag+2I- -->2Ag(s)+I2
0.200*(50/1000)=0.01
0.01/2 *2=0.01
0.01/0.245=

40.8mL

But the answer is 81.6mL, which part is wrong??
ty!
 
Last edited:

Answers and Replies

  • #2
867
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2I- -->2e+2I2

That 2 doesn't belong.
 
  • #3
Suy
102
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2I- -->2e+2I2

That 2 doesn't belong.

yes, but the answer still wrong though...
 
  • #4
867
0
0.01/2 *2=0.01
I don' see why you're dividing by 2 above.

Show your units with the numbers and it'll be easier to find your mistake.
 
  • #5
Suy
102
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0.01/2 *2=0.01
I don' see why you're dividing by 2 above.

Show your units with the numbers and it'll be easier to find your mistake.

2Ag+2e-->2Ag(s)
2I- -->2e+I2

0.01mol / 2mol I-(aq) * 2mol Ag(aq)
=0.01mol

0.01mol / 0.245mol/L
=0.0408L
=40.8mL

also, if i use 2AgCl + ZnI2 ---> 2AgI + ZnCl2
i can get 81.6mL, but this question is in my redox unit....
or is this a trick question?
thx
 
Last edited:
  • #6
867
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I think the error is here: 0.200*(50/1000)=0.01
The 0.200 is the molarity of ZnI2, but the molarity of the I- ions is actually twice that much since there are 2I- in that compound. So you should use a molarity of 0.400 for the I-.

You could use the net ionic/half-reaction equation 2Ag+ + 2I- → 2Ag + I2 (or the actual equation which I believe is 2AgCl + ZnI2 → 2Ag + ZnCl2 + I2) and use conversion factors to find the answer; I got the correct answer doing it this last way.
 
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  • #7
Suy
102
0
But it's doesn't make sense to me if the I- ions s twice
I- -->1e+ (1/2)I2
then 0.01mol will be the same, since
Ag+1e-->1Ag(s)
I- -->1e+ (1/2)I2
0.01mol/ 0.245mol/L,
or is the textbook answer is wrong??
 
  • #8
867
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[tex]\frac{0.200 ~mol~ ZnI_2}{1~ L~ZnI_2~solution} \rightarrow \frac{0.400~mol~I^-}{1~L~ZnI_2~solution}[/tex]

[tex]\frac{0.245~mol~AgCl}{1~L~AgCl~solution} \rightarrow \frac{0.245~mol~Ag^+}{1~L~AgCl~solution}[/tex]

Using 2Ag+ + 2I- → 2Ag + I2,
[tex]0.0500~L~ZnI_2~solution\left(\frac{0.400~mol~I^-}{1~L~ZnI_2~solution}\right)\left(\frac{2~mol~Ag^+}{2~mol~I^-}\right)\left({\frac{1000~mL~AgCl~solution}{0.245~mol~Ag^+}\right) = ~?[/tex]

This was how I worked it; perhaps you can see where you went wrong. I still can't see the problem exactly because you haven't shown the units on 0.01 mol.
 
Last edited:

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