another gen chem practice question

[tlcmqn]

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When the following reaction is balanced, what is the sum of the ionic charges on the right side of the equation?

(H+) + (MnO4 -) + (Fe 2+) ---> (Mn 2+) + (Fe 3+) + H2O

answer is 17+

I got 5+ after balancing the equation
this is how I balanced it

(4e-) + 8(H+) + (MnO4 -) + (Fe 2+) ---> (Mn 2+) + (Fe 3+) + 4(H2O)

I added the 2+ on Mn and 3+ on Fe and got an answer of 5+

What did I do wrong?

Again, the correct answer is 17+

 

[BaylorDDS]

you either balanced it wrong, or don't know what they mean by ionic charges (charges on atoms, not electrons)

So you balance it by seperating into half-rxns:
5e- + 8H+ + MnO4- --> Mn2+ + 4H20
Fe2+ --> Fe3+ + 1e-

Now you must multiply the entire bottom one (Fe one) by 5 to cancel the electrons. Giving you:
5e- + 8H+ + MnO4- --> Mn2+ + 4H20
5Fe2+ --> 5Fe3+ + 5e-

Now combine them:
5e- + 5Fe2+ + 8H+ + MnO4- --> Mn2+ + 4H20 + 5Fe3+ + 5e-

Now cancel everything out:
5Fe2+ + 8H+ + MnO4- --> Mn2+ + 4H20 + 5Fe3+

You have two ions: Mn2+ & Fe3+. So 5X3+ for Fe = 15. 15 + 2 for Mn2+ = 17

If you have any questions about balancing or whatever, feel free to pm me.