Standard Reduction potentials (1 Viewer)

[Dragie]

Ok...I was away when my class was doing the whole Standard reduction potential redox thing and I really have no idea what it is or what it does. Could someone explain how the E0 value comes into relevance for these half equations as well. Lol I need a lot of explaining here.
Thank you so much in advance!

 

[Dreamerish*~]

First, look at a Standard Reduction Potential Data Sheet.

Say you have a Galvanic cell with zinc as the anode and copper as the cathode, and you want to find the EMF (Electromotive force) of the cell.

The half equation for the anode is:

Zn → Zn²⁺ + 2e^-

Zinc is being oxidised. Now if you look at the reduction equation in the data sheet above, the E[FONT=&quot]° is for the reduction of zinc, that is:

Zn²⁺ [/FONT]+ 2e^- → Zn

The value shown is -0.76. However, in our Galvanic cell, zinc is oxidised, which is the opposite of reduced, so the E[FONT=&quot]° is the negative of the value shown. That is, +0.76.

Now for copper, which is reduced, the half equation is:

Cu²⁺ + 2e^- [/FONT]→ Cu

From the data sheet above, we can see that the E[FONT=&quot]° for this half equstion is 0.34.

To calculate the total [/FONT]E[FONT=&quot]MF of the cell, we add the two together. 0.76 + 0.34 = 1.1 V.[/FONT][FONT=&quot][/FONT][FONT=&quot][/FONT][FONT=&quot][/FONT][FONT=&quot][/FONT]

 

[Riviet]

In each of those half equations, the → represents an arrow pointing to the right, ie one of these → arrows.

P.S Dreamerish - to make those arrows display, don't bold them.

 

[Dreamerish*~]

In each of those half equations, the → represents an arrow pointing to the right, ie one of these → arrows.

P.S Dreamerish - to make those arrows display, don't bold them.

Ooh, thanks.

 

[richz]

remeber this,

The table shows reduction so whenever the reduction eqn is asked just write the eqn from the table but the oxidation eqn is the opposite way and so the voltage should also be reveresed.

Also to calculate the total emf just add both of them and u get the total emf.

Note: oxidation occurs on the anode
reduction occurs on the cathode.

the more reactive metal replaces the less reactive. The metal closer to lithium is the more reactive one. The more reactive metal usually gives away its electrons easily while the less reactive accepts electrons.

ie. So if iron is placed in Cu2SO4 soln, red metal is deposited on the iron. This is becuz iron is the more active metal and so easily gives away electrons so the Fe(s) -> Fe2+ but the copper ion changes to a solid: Cu2+ -> Cu (s) so a solid is formed therefore the iron is covered in red metal and the soln decolourises. iron replacing Cu in the soln turning it into FeSO4.


PS. Dreamerish where did u get this table from, it looks different to the normal (HSC) ones. Must have got it from some advanced chem class u went to .

 

[Dreamerish*~]

Dreamerish where did u get this table from, it looks different to the normal (HSC) ones. Must have got it from some advanced chem class u went to .

Googled it.

 

[Dragie]

Woohoo spank u