Combining half cells Flashcards
1
Q
WHat will happens if the voltmeter is released with a wire>
A
- electrons will flow from the area of many electrons to the area with few
- -ve electrode (negative Eϴto the positive)
2
Q
What happens to the position of the equilibrium when the electrons flow?
A
- according to Le Chatelier’s principle…
- (most) -ve electrode
- electrons lost
- position will move to left to replace lost electrons
- (most) +ve electrode
- number of electrons increases
- equilbrium moves to the right
- leads to reduction
3
Q
write the equation for this reaction
A
Cu2+(aq) + 2e- + Zn(s) → Cu (s) + Zn2+(aq) + 2e-
electrons cancel
4
Q
What is the rule of thumb for predicting the direction of electrochemical cells?
A
- most -ve Eϴ moves to the left
- most +ve Eϴ moves to right
- then combine using them directions
5
Q
Why is the reaction between copper and sulphuric acid not feasible?
A
- copper eqm already to right
- to react eqm would have to move to the left
- opposite to what Eϴ values are saying
- H+ would have to move to right
- against what values demand
6
Q
How do you choose an oxidising agent?
A
- recap
- oxd is loss of e-
- oxd agent oxidises something by removing electrons from it, gains electrons (itself is reduced)
- to encourage reduction
- eqm must be moved to left
- must couple it to a second eqm
- in order for reduction to occur, the second eqm Eϴ value must be greater
- removes electrons from the first eqm, shifts eqm to left
7
Q
How do you choose a reducing agent?
A
- recap
- reduction is gain
- reducing agent gives electrons, itself is oxidised
- substance losing electrons is found on the right-hand side of one of these redox equilibria
- to reduce something, it must be couples with a second quilibrium with a more negative value
- substance w/ most -ve value acts as a reducing agent
- electrons lost and moved into the first eqm mixture
- Le chateliers principle
- eqm moves to right
