# Ionic equation

• June 8th 2011, 07:46 AM
igcsestudent
Ionic equation
Hi guys,
I was doing this past paper, but I got the ionic equation wrong, it is (v) in the attached file, it should be Fe3+, which means that it lost one electron, I know that is had been oxidized, but since the bromine gained two electrons, shouldn't the iron lose two as well to become Fe4+?. Can anyone please explain this to me?
• June 8th 2011, 09:09 AM
mrfour44
The cell shown in your PDF document is a Galvanic cell. It converts chemical to electrical energy by means of redox reactions and yes, electron transfer. The inert electrodes are named according to their environments and their nature themselves. The right (-) electrode is the anode because iron has a greater electrode potential (ability to oxidize). Analogously, the left (+) cathode is the cathode. The electrode with a better oxidizing ability is the one in whose vicinity oxidation occurs.

If we take A to be the anode and C the cathode, then:

$A(-): \mathrm{Fe^{2+} (aq)}\rightarrow\mathrm{Fe^{3+} (aq) + 1\,e^-}$
$C(+): \mathrm{Br_2 (aq) + 2\,e^-}\rightarrow\mathrm{2\,Br^-(aq)}$

The rule of thumb with reduction-oxidation reactions is: A substance can be reduced if and only if another substance is oxidized. Therefore, to answer your question, YES!, two electrons are released, but, from two respective bivalent iron $\mathrm{Fe^{2+}}$ cations. Therefore we find the "common denominator" and balance the amount of electrons released and absorbed.

$A(-): \mathrm{2\,Fe^{2+} (aq)}\rightarrow\mathrm{2\,Fe^{3+} (aq) + 2\,e^-}$
$C(+): \mathrm{Br_2 (aq) + 2\,e^-}\rightarrow\mathrm{2\,Br^-(aq)}$

Then, in the sum equation, the electrons cancel each other out:

$\mathrm{2\,Fe^{2+} (aq) + Br_2 (aq)\rightarrow 2\,Fe^{3+}(aq) + 2\,Br^-(aq)}$

Hope I helped!

Just remember: A substance can reduce only if it has the electrons to do so. You should always check if your stoichiometric coefficients are in alignment.

EDIT: By the way! Iron can never exist as $\mathrm{Fe^{4+}}$ because the ionization energy to expel a fourth electron (one that is not in its valence shell) is immeasurably high. So iron will always exist as either $\mathrm{Fe^{2+}}$, with one valence electron, or $\mathrm{Fe^{3+}}$, without any at all.
• June 8th 2011, 09:25 AM
igcsestudent
Thanks
I really appreciate you taking your time to help :-)
• June 8th 2011, 09:26 AM
mrfour44
Happy to help! Just remember the rule of thumb, it saved me more times than I could count. !
• June 8th 2011, 09:32 AM
igcsestudent
Quote:

Originally Posted by mrfour44
The cell shown in your PDF document is a Galvanic cell. It converts chemical to electrical energy by means of redox reactions and yes, electron transfer. The inert electrodes are named according to their environments and their nature themselves. The right (-) electrode is the anode because iron has a greater electrode potential (ability to oxidize). Analogously, the left (+) cathode is the cathode. The electrode with a better oxidizing ability is the one in whose vicinity oxidation occurs.

If we take A to be the anode and C the cathode, then:

$A(-): \mathrm{Fe^{2+} (aq)}\rightarrow\mathrm{Fe^{3+} (aq) + 1\,e^-}$
$C(+): \mathrm{Br_2 (aq) + 2\,e^-}\rightarrow\mathrm{2\,Br^-(aq)}$
.

I don't mean to be rude or criticize you.... but I think you did a small mistake, the anode is positive, remember, PA, positive anode...
I hope I helped you just like you helped me :-)
• June 8th 2011, 10:08 AM
mrfour44
No worries.

In an electrolytic cell the anode is the positive electrode, yes, but in a Galvanic cell the anode is the negative electrode because of the more negative electrode potential (ability to oxidize).

See: Anode - Wikipedia, the free encyclopedia and Electrode potential - Wikipedia, the free encyclopedia

Btw, I do correct myself - iron has a more negative electrode potential, and that's why it's at the anode.
• June 8th 2011, 10:19 AM
igcsestudent
Sorry, my chemistry is very limited...
Thanks for all the help