Question

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Pb2+ + 2e- → Pb -0.13 V

How would the following changes alter the potential of the cell?

Adding Ag+ ions to the silver half reaction (assume no volume change)

Adding Pb2+ ions to the lead half reaction (assuming no volume change).

Removing Pb2+ ions from solution by precipitating them out of the lead half reaction (assume no volume change). Adding equal amounts of water to both half reactions.

increase, decrease or no change in potential

Answer 1

Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.

Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V

All other samples are based on this reference.

Find the Reduction Potential of each reaction (Tables)

Ag+ + e- → Ag 0.80 V

Pb2+ + 2e- → Pb -0.13 V

The most positive has more potential to reduce, it will be reduced

The most negative will be oxidized, since it will donate it selectrons

For total E°cell potential:

E°cell = Ered – Eox

Eox = -Ered of the one being oxidized

E°cell = 0.80- (-0.13) = 0.93 V

E°cell = 0.93 V

a)

More Ag+, reactants, will shift toward MORE products, expect more electron flow, thereofre, increase in electron potential

b)

addition of Pb+2, a product, can't be used for more electorn flow, so this will decrease the potential

c)

removal of PB+2 will favour the potential

d)

addition of water will: DECREASE concentrations

This affects most to Q

Q = [Pb+2] / [Ag+]^2

as Q increases, this favours LEAST flow of electorns, that is, lower potential