Cathodic protection (1 Viewer)

[mushroom_head]

Can someone please tell me how cathodic protection works. I'm really confuused.

 

[Xayma]

A more reactive metal is placed on the cathode.

If you do the sums, then you will see the oxidation of the metal and reduction of the created metal ions that you are trying to protect is positive ie it is spontaneous.

When the metal gives up its electrons the sacraficial anode oxidises and gives its electrons to the metallic ions.

 

[mushroom_head]

ok..thanx Xayma.

You know in a galvanic cell, would the more reactive metal be placed on the anode?

 

[Tommy_Lamp]

well you cant exactly place something on the anode
the more reactive metal BECOMES the anode

 

[~*HSC 4 life*~]

ok..thanx Xayma.

You know in a galvanic cell, would the more reactive metal be placed on the anode?

yes, anything more reactiive has a higher potential for oxidation, and oxidation occurs at the anod in glavanic cells

 

[steph@nie]

the more reactive metal would be the anode

Cathodic protection is used to protect ships from rust. there's two types- impressed current and sacrificial anode.
Impressed current is where a circuit is made with the ship or pipe or whatever needs protecting, at the cathode and an inert anode. A small voltage is applied, with water completing the circuit. O2 is reduced at the cathode and water is oxidised at the anode.
Sacrifical anode involves blobs of a metal such as Zn being attached to the ship. The blobs become the anode, and are oxidised or 'sacrificed'.

 

[bobbie212]

with oxidation must you also have reduction?
This is what I dont get in cathodic protection, where do the metalic ions of the least reactive metal say iron come from? If it is from the oxidation of the iron, that means we have oxidation of the more reactive metal (say zinc), the oxidation of iron and then the reduction of iron ions. So we are missing another reduction?

please some1 clear that up cheers

 

[Xayma]

The reduction reaction that generates the iron ions will most often be:

1/2 O_2(g)+H₂O_(l)+2e^- ------>2OH^-

 

[Tommy_Lamp]

then OH- + Fe2+ = Fe(OH)2

right?

 

[mushroom_head]

by the way, you know when you place the electrode in a concentrated electrolyte, the water doesn't take much part in the reaction right? So in the examn if they ask you to write down the oxidation and reduction equations, do you just completely ignore H2O even if it takes less energy to react than the other substance?

 

[Xayma]

then OH- + Fe2+ = Fe(OH)2

right?

Then 4Fe(OH)₂+O₂ ---> 2(Fe₂O₃.xH₂O)+(2-x)H₂O

Basically in a concentrated electrode the only substance that will change if water requires less energy is the chloride ions may be oxidised.

Water will still be reduced in favour of K⁺, Na⁺ etc.
And Water will still be oxidised in favour of F^-