✔ 最佳答案
0.
Mind your formulae. They're "MgSO4", "CuSO4", KNO3" etc.
1.
First of all, NO. There's NO reaction between Mg(2+) and NO3(-), or btw K(+) and SO4(2-).
If they form precipitate, then there're reactions; but in these cases, they still exist as mobile ions when they meet.
(In fact, there're a lot of ways to change the composition of solutions w/o chemical rxn.)
Then, the SO4(2-) ions are naturally present with Mg(2+) - when you dissolve MgSO4 in water to give your electrolyte solution. Or you may say, SO4(2-) exists to balance the electrical charges (of the magnesium ions).
And, yes, Mg(2+) may move to another half-cell through diffusion. But that would be slow and insignificant, so don't bother with it within normal experimental conditions.
2.
Still wrong grammar. **Does it mean that**
There're several causes for rxn. to stop:
- exhaustion of reactants (your electrolytes, electrode materials, and sometimes acids / bases for some reactions)
- the half-reactions attain equilibrium
I'll elaborate on 2nd reason.
Cu(2+) + 2e- <====> Cu
Mg <====> Mg(2+) + 2e-
In first reaction, copper ions are being used up.
Conc. of Cu(2+) ion decreases, so eqm. position lies on LHS.
That means, this half-cell becomes less and less likely to accept electrons.
In second reaction, magnesium ions are being produced.
Conc. of Mg(2+) ion increases, so eqm. position lies on LHS.
That means, this half-cell becomes less and less likely to give out electrons and Mg(2+) ions.
As reaction proceeds, it'll eventually reach a point that the Cu-half cell takes no more electrons, and Mg-half cell gives no more electrons.
Even NOT all the reactants are used up, reaction still stops (reactants have no more "motivation" to do the reaction).
You can learn more on this by looking at "standard reduction potential" and "Nernst equation".
