module 3 chem (1 Viewer)

[kractus]

When the ionisation of a metal is measured, the equation is as follows:
M(x) + energy --> M(y) + e-
What are the states of x and y respectively?
(A) g, g
(B) s, g
(C) s, s
(D) g, s

could someone explain why its A?

 

[kractus]

Also, what would be the balanced half equation at the cathode?

 

[wizzkids]

Question 1.
In order to measure the energy to separate one electron from an atom, you have to eliminate all the other variables that could influence the energy of the original neutral atom and the ions that are produced. That means there can't be any solvents, or any metallic bonds, or or any chemical bonds of any sort. That means the starting point has to be an isolated atom, which is another way of saying it must a gas. The ions also have to be isolated, so they are gas as well.
Does that make sense?
Question 2.
I note that you have filled the right-hand cell with ferric nitrate. There are two possible reduction reactions at the cathode. The first is reduction of ferric (Fe III) ion to ferrous (Fe II) ion and then when most of the Fe(III) ions are gone you will get reduction of Fe(II) to Fe(s).
Do you think you can write the cathode half-equation now?

 

[kractus]

Question 1.
In order to measure the energy to separate one electron from an atom, you have to eliminate all the other variables that could influence the energy of the original neutral atom and the ions that are produced. That means there can't be any solvents, or any metallic bonds, or or any chemical bonds of any sort. That means the starting point has to be an isolated atom, which is another way of saying it must a gas. The ions also have to be isolated, so they are gas as well.
Does that make sense?
Question 2.
I note that you have filled the right-hand cell with ferric nitrate. There are two possible reduction reactions at the cathode. The first is reduction of ferric (Fe III) ion to ferrous (Fe II) ion and then when most of the Fe(III) ions are gone you will get reduction of Fe(II) to Fe(s).
Do you think you can write the cathode half-equation now?

thank u would it be the fe3+/fe2+?

Also, another question:
Metal M will displace copper from copper sulphate solution, silver from silver nitrate solution and tin in a solution of tin (II) chloride. However, metal M will not displace zinc form the zinc chloride solution. The most likely value for the E of the M/M(n+) half-cell is?

 

[wizzkids]

Metals Activity Series.
A certain metal (solid) will displace Cu²⁺ ions, Ag⁺ and Sn²⁺ ions. It does not displace Zn²⁺ ions. Therefore we look up the activity series of metals, and we discover that metal M is probably iron or nickel, because only these two metals have a standard reduction potential lying between those of tin and zinc.
Can you work out what the standard reduction potential E^o value will be ?