Why is there an H+ there? Is it just to show that the MnO4- is acidifed?
Same with this one:
Cu + H+ + NO3- ---> Cu2+ + NO2
Thanks for any help!
H^+ is there to show that the solution is acidified, but also because it is required to convert the oxygens on the anions into water, which is missing from the right hand sides of both of those equations.
H^+ is there to show that the solution is acidified, but also because it is required to convert the oxygens on the anions into water, which is missing from the right hand sides of both of those equations.
Now try balancing and see how far you get.
Okay, I understand. So if I worked it out right, I think these would be the answers for both the equations:
Okay, I understand. So if I worked it out right, I think these would be the answers for both the equations:
16H+ + 2MnO4- + 10I- ---> 5I2 + 2Mn2+ + 8H2O
4H+ + 2HNO3- + Cu ---> Cu2+ + 2NO2 + 2H2O
I just worked them out how I usuallt do I guess.
I do believe those are correct.
You can use the method of trial and error, you can work out what the half equations are and derive the equation from those, or you can use oxidation states to establish the reacting ratios.
EDIT: the second equation should read as 4H+ + 2NO3- + Cu ---> Cu2+ + 2NO2 + 2H2O, but other than that, your answers are perfect
You can use the method of trial and error, you can work out what the half equations are and derive the equation from those, or you can use oxidation states to establish the reacting ratios.
Thank you very much! Yeah I used the half equation method. Works really well for me. I was just really confused why there was an H+ as all the questions I'm used to don't have it lol.
Thanks again for the support and I hope you have a good night!