Question

need help

Answer 1

for #1 have 5Fe^2+(aq)+MnO4-+8H(aq) am i correct?

Please show me how to do #2?

Answer 2

Sorry, I don't recall redox..

Answer 3

I am sorry, I don't know chemistry. hihihi... I can balance the available equation but not know how to write it out.

Answer 4

Sorry, I can't help with chemistry

Answer 5

okay thank anyway

Answer 6

For the first one, 5Fe^2+ + MnO4- + 8H --> 4H2O + 5Fe3+ + Mn2+
Don't forget the products!

For the second one, do you have a balanced equation that you want me check? Or do you want me to go through it with you?

Answer 7

that ne i need hep with cause idk what t d0

Answer 8

@matt101 h0w w0uld i d0 the sec0nd 0ne

Answer 9

First let's come up with our balanced half reactions. Let's start with chromium because it's a bit harder to balance. Once we've found this half reaction, we can easily modify the iron half reaction to fit what we have:

\[Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow 2Cr^{3+}\]
Cr3+ is a product that was left out of your equation above. You can see the oxidation state of Cr is +6 on the reactant side, but becomes +3 on the product side. This means each chromate ion must have gained 3 electrons, meaning 6 electrons were added overall. Now we need to deal with the oxygen. This is happening in an acidic solution, so let's add 7H2O to the products to balance the oxygen atoms on the reactant side:

\[Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow 2Cr^{3+}+7H_{2}O\]
However, by adding water, we've added a bunch of hydrogen atoms into the mix as well. Good thing we're in an acidic solution! Let's add 14 H+ ions to the reactant side to balance those:

\[Cr_{2}O_{7}^{2-}+6e^{-}+14H^+\rightarrow 2Cr^{3+}+7H_{2}O\]
Now this half reaction is balanced! Now onto the iron half reaction. This one's easy to balance easy to balance because we can clearly see we're just losing an electron:

\[Fe^{2+} \rightarrow Fe^{3+} +e^-\]
However, look at our chromium half reaction - there are 6 electrons there! electrons are not part of our final reaction, so they must cancel each other. Let's multiply the iron half reaction by 6 to make this happen:

\[6Fe^{2+} \rightarrow 6Fe^{3+} +6e^-\]
Now we can add the two half reactions to come up with our full reaction:

\[6Fe^{2+} +Cr_{2}O_{7}^{2-}+6e^{-}+14H^+\rightarrow 6Fe^{3+} +6e^-+2Cr^{3+}+7H_{2}O\]\[6Fe^{2+} +Cr_{2}O_{7}^{2-}+14H^+\rightarrow 6Fe^{3+} +2Cr^{3+}+7H_{2}O\]

And that's it! Hope that helps!

Answer 10

thanks for explaining this clearly in a way i can understand.