Hi ya'll,
I have the following problem:
Cr2O7^-2 + H^+ + I2 ---> Cr^+3 + H2O + IO^-
===========================================
Cr2O7^-2 + H^+ + I2 ---> Cr^+3 + H2O + IO^-
Split Redox #1:
Cr2O7^-2 + 14H^+ + 3e^- ---> Cr^+3 + 7H2O
2[Cr2O7^-2 + 14H^+ + 3e^- ---> Cr^+3 + 7H2O]
2Cr2O7^-2 + 28H^+ + 6e^- ---> 2Cr^+3 + 14H2O
Split Redox #2:
I2 + 2H2O + 2e^- ---> 2IO^- + 4H^+
3[I2 + 2H2O + 2e^- ---> 2IO^- + 4H^+]
3I2 + 6H2O + 6e^- ---> 6IO^- + 12H^+
Split Redox #3:
H^+ + OH^- ---> H2O
Overall:
3I2 + 6H2O + 6e^- ---> 6IO^- + 12H^+
2Cr2O7^-2 + 28H^+ + 6e^- ---> 2Cr^+3 + 14H2O
H^+ + OH^- ---> H2O
My question is: How do cancel out the electrons and hydrogen cations? This is all new to me so cut me some slack for going this far. I am still learning and I love redox, but this problem is giving me a hard time. I would like a chemist major or anyone to help out on this. Thanks in advance!! :)
Key: "^" is the charge raised so I won't confused anyone reading this problem.
Help with: Redox Reaction Dilemma?
2013-12-07 7:59 pm
回答 (2)
2013-12-07 9:58 pm
✔ 最佳答案
1st half eq'n Cr2O7^2- + 14H^+ + 6e^- = 2Cr^3+ + 7H2O
2nd half eqn'
I2 + 2H2O + 2e^- ---> 2IO^- + 4H^+
Rewriting to balance the charges.
I2 + 2H2O = 2IO^- + 4H^+ + 2e^-
Multiply the 2nd half eq'n by '3'.
Hence
3I2 + 6H2O = 6IO^- + 12H^+ + 6e^-
Cr2O7^2- + 14H^+ + 6e^- = 2Cr^3+ + 7H2O
Add the two half eq'ns to eliminate the 'e^-'
Cr2O7^2- + 14H^+ + 3I2 + 6H2O = 2Cr^3+ + + 7H2O + 6IO^- + 12H^+
We now balance the water and the acids by subtracting the corresponding values from both sides.
Hence
Cr2O7^2- + 2H^+ 3I2 = 2Cr^3+ + H2O + 6IO^-
The fully balanced Redox eq'n.
It is an algebraic process.
Hope that helps!!!!
2013-12-08 5:00 am
Cr2O7^-2 + 14H^+ + 3e^- ---> Cr^+3 + 7H2O is WRONG. (Cr and charge on each side 9+ & 3+ not balanced
It should be
Cr2O7^-2 + 14H^+ + 6e^- ---> 2Cr^+3 + 7H2O (2Cr and 6+ on each side)
3I2 + 6H2O + 6e^- ---> 6IO^- + 12H^+ is WRONG (6e are on wrong side, makes both your equations reductions!, charge is 6- on left, 6+ on right
It should be
3I2 + 6H2O ---> 6IO^- + 12H^+ + 6e^- ( 0 charge on each side)
Now add together my two half equations and e's will be eliminated, also collect H2O onto one side
Cr2O7^-2 + 14H^+ + 6e^- + 3I2 + 6H2O ---> 2Cr^+3 + 7H2O + 6IO^- + 12H^+ + 6e^-
Cr2O7^-2 + 14H^+ + 3I2 ---> 2Cr^+3 + H2O + 6IO^- + 12H^+
As well as all the atoms being balanced 2Cr, 7 O, 14H, 6I it is important that charge is the same on each side 12+
It should be
Cr2O7^-2 + 14H^+ + 6e^- ---> 2Cr^+3 + 7H2O (2Cr and 6+ on each side)
3I2 + 6H2O + 6e^- ---> 6IO^- + 12H^+ is WRONG (6e are on wrong side, makes both your equations reductions!, charge is 6- on left, 6+ on right
It should be
3I2 + 6H2O ---> 6IO^- + 12H^+ + 6e^- ( 0 charge on each side)
Now add together my two half equations and e's will be eliminated, also collect H2O onto one side
Cr2O7^-2 + 14H^+ + 6e^- + 3I2 + 6H2O ---> 2Cr^+3 + 7H2O + 6IO^- + 12H^+ + 6e^-
Cr2O7^-2 + 14H^+ + 3I2 ---> 2Cr^+3 + H2O + 6IO^- + 12H^+
As well as all the atoms being balanced 2Cr, 7 O, 14H, 6I it is important that charge is the same on each side 12+
收錄日期: 2021-04-20 20:32:50
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