[ponyoponyo2]

Are these answers correct?

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Consider the following Galvanic cell at 291 Kelvin:

Cu|Cu2+||Au3+|Au

where the standard cell potential is 1.18 V. How will the following changes affect the potential of the cell? (Increase, Decrease, or No Change)

a. Increase the mass of the Au
b. Adding equal amounts of H2O to both the anode and cathode
c. Removing H2O from the anode

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I know the balanced equation is:

3Cu + 2Au³⁺ --> 3Cu²⁺ + 2Au

a. No change since Au is a solid
b.  Increase since the coefficient for Cu²⁺ is higher than that for Au³⁺, Cu²⁺ would be more diluted
c. Increase because the Cu²⁺ is concentrated

[Vidya]

here K=[Cu2+]^2/[Au3+]^3
solid is not the part of K and adding / removing water will change the concentrations

[ponyoponyo2]

Thanks but you didn't really answer my question...

[DevaDevil]

here K=[Cu2+]^2/[Au3+]^3
solid is not the part of K and adding / removing water will change the concentrations

I assume you mean K = [Cu²⁺]³ / [Au³⁺]²

Thanks but you didn't really answer my question...

he did give you enough to be able to reason with le Chateliers principle, but for electrochemical cells it is best to remember the nernst equation, which relates the equilibrium constant and the cell potential:

E_{half cell} = E_{half cell}⁰ - (RT/nF) ln (a_reductant/a_oxidant)

concentrations are usually substituted for activity, and activity of solids is 1.
Definition: Oxidant + n e^- <--> Reductant

so four your system:

E_Au = E_Au⁰ - (RT/6F) ln 1/[Au³⁺]²
E_Cu = E_Cu⁰ - (RT/6F) ln 1/[Cu²⁺]³
attention: already normalized to 6-electrons

E_cell = E_Au - E_Cu
E_cell = E_Au⁰ + (RT/6F) ln [Au³⁺]² - (E_Cu⁰ + (RT/6F) ln [Cu²⁺]³ )
E_cell = (E_Au⁰ - E_Cu⁰ ) + ln [Au³⁺]² ) - (RT/6F) (ln [Cu²⁺]³
E_cell = (E_Au⁰ - E_Cu⁰ ) +  (RT/6F) ln ([Au³⁺]²/[Cu²⁺]³ )


Or: E = E⁰ - (RT/nF) ln K, as it is more generally written



And to check your answers:
1. correct
2. correct
3. Incorrect