Briefly comment on the effect of solution concentration on cell potential. Note

Question

Briefly comment on the effect of solution concentration on cell potential. Note that E degree is measured using 1M solution, while your measurements were made using 0.1M solutions. A student doing this experiment using a U-tube salt bridge is having difficulty obtaining useful voltage readings. Examining the experimental setup, the student finds an air pocket between the stopper at one end of the U-tube and the electrolyte solution inside the tube. Briefly explain how the air pocket could be the source of the erroneous voltage readings. If the cell you constructed in this experiment were doubled in size, would the measured cell potentials double? Briefly explain.

Explanation / Answer

1.

The effect of solution concentration in cell potential

Recall that

Estd = the standard value of the cell potential of a cell. This includes: T = 25C and concentrations of ALL species at C = 1M

that is, this is a very unique case. In real life we will have different concentrations of species

This is why we have this questions... What will be the effect of concentration in the Ecell value?

Let's start intuitively before jumping into equations

recall that:

the more reactants we have, the more products we will form in the future

the less reactatns we have, the less products we will form in the future

therefore, the more reactants --> more reaction --> more flow of electrons

more flow of electrons --> more electricity or Electromotive force

then

we will want to have HIGH amount of reactants

This is stated as follow sin the Nernst Equation

Ecell = Estd - 0.0592/n * log(Q)

where:

Ecell = the final value of the cell

Estd = standard Ecell value (at standard conditions)

n = number of electrons being transferred

Q = the coeffiecient between products and reactants

therefore... sicne n is constant for any reaciton (i.e. the reaction will still have the same electrons flowing since it does not depends on the amount of material present)

Note htat Estd is also constant, since std conditions are fixed, that is, T = 25 C and C = 1 M

then

only Q is changing

Q is given as

Q = [products]^P / [reactants]^R

note that if products are large, then Q will be large

if Reactants are larger than products, Q will be very small

since we are using LOG operation, we do need very low amount of products and high amount of reactants to have a negative value of that log(X) value

then, negative * negative = positive

Ecell turns to

Ecell = Estd + Y-value

then

Ecell > Estd; as long as Q < 1

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