Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Glitch000 on November 17, 2021, 09:16:43 AM
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hi, i can't resolve this redox reaction, can someone help to find the error?
P4O6 + I2 = P2I4 + P4O10
first i found the oxidation number and then i wrote the semi reaction and I balanced them:
Red1: 12H+ + 4e- + P4O6 = 4P+2 + 6H2O
Red2: 4e- + 2I2 = 4I-
Then i added them toghether:
12H+ + 8e- + P4O6 + 2I2 = 4P+2 + 6H2O + 4I-
Ox: 4H2O + P4O6 = P4O10 + 8e- + 8H+
The total arrives:
12H+ + 8e- + P4O6 + 2I2 + 4H2O + P4O6 = 4P+2 + 6H2O + 4I- + P4O10 + 8e- +8H+
at the end my balanced reaction arrives:
2P4O6 + 2I2 + 4H+ = 4P2I4 + P4O10 + 2H2O
The reaction is wrong because the exact balanced reaction is this:
5P4O6 + 8I2 = 4P2I4 + 3P4O10
I don't understand were is my error... ???
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Wow. Just wow. I would never think it is possible to overcomplicate thing by that much. Try to balance just by inspection (most likely starting with iodine will be the easiest approach).
But:
Red1: 12H+ + 4e- + P4O6 = 4P+2 + 6H2O
Red2: 4e- + 2I2 = 4I-
Then i added them toghether:
12H+ + 8e- + P4O6 + 2I2 = 4P+2 + 6H2O + 4I-
This is already wrong - when you add these two reactions you assume they occur in some particular ratio. Why?
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at the end my balanced reaction arrives:
2 P4O6 + 2 I2 + 4 H+ = 4 P2I4 + P4O10 + 2 H2O
Of course its wrong. Did you ever read what have you written here ?
Where is the error:
Red1: 12 H+ + 4 e- + P4O6 = 4 P2+ + 6 H2O
Red2: 4 e- + 2 I2 = 4 I-
Then i added them toghether:
12 H+ + 8 e- + P4O6 + 2 I2 = 4 P2+ + 6 H2O + 4 I-
Formula of the compound is the P2I4, not P2I22+ as you have written (4 P2+ + 4 I-) (Red2 needs to be multiplied by 2).