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Redox stoichiometry

  1. Oct 24, 2009 #1

    Suy

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    1. The problem statement, all variables and given/known data

    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)
    2. Relevant equations



    3. 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: Oct 24, 2009
  2. jcsd
  3. Oct 24, 2009 #2
    2I- -->2e+2I2

    That 2 doesn't belong.
     
  4. Oct 24, 2009 #3

    Suy

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    yes, but the answer still wrong though...
     
  5. Oct 24, 2009 #4
    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.
     
  6. Oct 24, 2009 #5

    Suy

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    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: Oct 24, 2009
  7. Oct 24, 2009 #6
    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.
     
    Last edited: Oct 24, 2009
  8. Oct 25, 2009 #7

    Suy

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    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??
     
  9. Oct 25, 2009 #8
    [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: Oct 25, 2009
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