Asked by dany
use the standard potential reduction potentials table to balance the following redox equation:
H2O2+I-+H+==>H2O+I2
would what i did be correct
H2H2+I^-+H^+==>H2O+I2
H2H2+H^+==>H2O
And I^-==>I2
Give us H2H2(aq)+2H^+(aq)+2e-==>2H2O(l)
And 2I^-(aq) ==> I2(s)+2e-
Add to getter and get:
H2H2(aq)+2H^-+2I^-(aq)==>2H2O(l)+I2(s)
H2O2+I-+H+==>H2O+I2
would what i did be correct
H2H2+I^-+H^+==>H2O+I2
H2H2+H^+==>H2O
And I^-==>I2
Give us H2H2(aq)+2H^+(aq)+2e-==>2H2O(l)
And 2I^-(aq) ==> I2(s)+2e-
Add to getter and get:
H2H2(aq)+2H^-+2I^-(aq)==>2H2O(l)+I2(s)
Answers
Answered by
DrBob222
Close but not quite.
H2O2 and not H2H2 but I suppose that is just a typo.
The H^+ on the left has a + charge and not a - charge. Final equation should be
2H^+ + 2I^- + H2O2 ==> I2 + 2H2O
H2O2 and not H2H2 but I suppose that is just a typo.
The H^+ on the left has a + charge and not a - charge. Final equation should be
2H^+ + 2I^- + H2O2 ==> I2 + 2H2O
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