I\'m doing ap chem electrochemistry chapter and i was wondering why dead battery

Question

I'm doing ap chem electrochemistry chapter and i was wondering why dead battery (reaching to 0 voltage) means it reaches at equilibrium. Can sombody explain this in detaill?
Why does too much reactant mean increase in E cell potential?? I saw one explanation that said that as there is more reacrant than product, the reaction more powerfully proceeds in order to reach equilibrium. I don 't really get this. Also, how can I say the standard cell potential (e°)is increasing or decreasing based on Q value (product over reactant)

PLEASE HELP MEE :(

Explanation / Answer

Note that "equilibrium" means that the reactants rate of reaction equals the proucts rate of reaction as well

so

A <-> B

the rate of A turning to B is the same as B turning to A

so... for a galvanic cell:

Assume:

Cu2+ + 2 e Cu(s) +0.337

Zn2+ + 2 e Zn(s) 0.7618

Cearly, the forward reaction is:

Cu2+ +Zn(s) Cu(s)+Zn2+ E = +0.337 + 0.7618 = 1.0988V

so..

this will not be "infinite" there will be a point in time in which this will stop...

Ecell = E°cell -0.0592/n*log(Q)

where Q is the product/reactant quotient

Q = [Zn+2]/[Cu+2]

Ecell = E°cell -0.0592/n*log(Q)

E°cell is constat, and n = electron transfer, the only thing that changes with time will be Concentrations, since reactants form product

So

Eventually

Ecell will turn up to 0

this is given when:

Ecell = E°cell -0.0592/n*log(Q)

so:

if Q <<< 1 the E°cell = 0.0592/n*log(Q)

meaning that Ecell = 0, so there will be no drive for reaction so this is in equilibrim!

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