Electrochemical Series question

[lauzy358]

Hello!

Quick question about the electrochemical series...could someone help me out please?

How do you know which half-equations are relevant regarding H₂O?

i.e.
In the electrochemical series (VCE), there's:
O_2(g) + 4H⁺_(aq) + 4e^- = 2H₂O_(l) [E⁰ = 1.23]
O_2(g) + H₂O_(l) + 4e^- = 4OH^-_(aq) [E⁰ = 0.40]
2H₂O_(l) + 2e^- = H_2(g) + 2OH^-_(aq) [E⁰ = -0.83]

From my (hopefully correct) understanding, if it's an aqueous solution the first one could be relevant.
What about the others...? =S
Say, if you have a solution of sodium hydroxide in a cell with etc. How do you know the reaction involved is the second option and not the third? Or can it be both and it comes down to competition? (but I doubt that)
I get the ones with hydrogen peroxide involved, as you'd only use those if the question involved hydrogen peroxide. Same goes for the hydrogen gas half-equation (I think). But this lot has me confused!

Thanks in advance (:

[Collin Li]

Water can act as an oxidant or a reductant, so both the top and bottom reactions are eligible. The middle one is unlikely unless you have oxygen gas specifically being pumped in at the electrode.

[dusty_girl1144]

The Edison cell is a 1.3 volt storage battery that
can be recharged, even after long periods of being
left uncharged. Its electrolyte is 21% potassium
hydroxide solution and the reaction on discharge
is:

Fe(s) + 2NiO(OH)(s) + 2H2O(l) ----> Fe(OH)2(s) + 2Ni(OH)2(s)

(a) Give electrode reactions during:
(ii) discharging
(iii) recharging

(b) What materials would be used for the
electrodes?


the answers for a are

(i) anode: 2H2O(l) + Fe(s) --> Fe(OH)2(s) + 2H+(aq) + 2e–

cathode: Ni(OH)3(s) + H+(aq) + e– ---> H2O(l) + Ni(OH)2(aq)

(ii) anode: H2O(l) + Ni(OH)2(aq) ---> Ni(OH)3(s) + H+(aq) + e–

cathode: Fe(OH)2(s) + 2H+(aq) + 2e– ---> 2H2O(l) + Fe(s)


the question im asking is how did they get these answers?
im really struggling with this topic, let alone equilibrium....
i really wanna get a good mark for this exam and for those who know what i got for my chem exam, u'll know how much i want this.

[Collin Li]

Determine the oxidation states of the species in the overall reaction to isolate which species are oxidising and reducing:

- Fe goes from an oxidation state of 0 to an oxidation state of +2 in iron (II) hydroxide. (oxidation)
- Nickel, in that compound goes from +3 to +2. (reduction)

Now, use the "KOHESE" method on each of those statements to obtain half-equations.

Example

To obtain this half-equation: "Fe goes from an oxidation state of 0 to an oxidation state of +2 in iron (II) hydroxide. (oxidation)"

Begin with:

The redox-active element (Fe) is already balanced, so go on to balance number of oxygens, by adding waters:



Now, balance the number of hydrogens, by adding protons:



Now, balance the charge, by adding electrons:



Repeat this process for this line: "Nickel, in that compound goes from +3 to +2. (reduction)" to get the other half-equation.

Hint: You'd begin with:

[dusty_girl1144]

....stupid question.... whats "KOHESE"

and also how do you know.... in discharging and recharging what the cathode and anode is? :s

[Collin Li]

....stupid question.... whats "KOHESE"

and also how do you know.... in discharging and recharging what the cathode and anode is? :s

Apparently it's supposed to be "KOHES" without the extra E (thanks caramel) but I never really use that acronym.

It's just the process of:

1) Balance the redox-active element (Fe, for example)
2) Balance oxygens (with water)
3) Balance hydrogens (with protons)
4) Balance charge (with electrons)

I described how you would write the half-equations for the discharging process (because I was analysing the change in oxidation states for the discharging reaction that they gave). The recharging process is the exact reverse, so just write your half-equations backwards, and those are your new reactions.

Since you've reversed your reactions, the oxidation reaction (anode during discharging) is now a reduction (hence the cathode while recharging), and the same swaparoonie happens with the reduction reaction too.

But to keep things simple, just remember the basics:

- Cathode is where reduction happens (gain of electrons)
- Anode is where oxidation happens (loss of electrons)

So just check whether the electrons are reactants (gain) or products (loss) in your new reversed half-equations to see which one is which.

[dusty_girl1144]

thanxs so much coblin! much appreciated. i understand it fully

[orsel]

Its electrolyte is 21% potassium hydroxide solution

It's just the process of:

1) Balance the redox-active element (Fe, for example)
2) Balance oxygens (with water)
3) Balance hydrogens (with protons)
4) Balance charge (with electrons)

Isn't that only for acidic solutions? I could be wrong though due to my infinite stupidity.

[Collin Li]

For alkaline solutions, you do the same thing, but there is an additional step:

Add hydroxides ions to both sides to neutralise the acid (which will become water). Cancel waters as necessary.

This was something I didn't learn until after VCE though, so if you ever have an unbalanced half-equation, you can feel free to use that method. I noticed that the method I suggested will not give you the answer that dusty_girl1144 posted, so that question isn't valid for VCE purposes.