Galvanic Cell

[ecoli]

Hi guys. I'm trying to figure out how to do this question. It was on a test at the beginning of the year, and now I have to know it for the final.

 

 

You have a galvanic cell based on the two half reaction:

 

[ce]Cr2O7^{-2} +14H^+ + 6e^- -> 2Cr^{+3} + 7H2O[/ce] [math]1.33V[/math]

[ce]Cu^{+2} + 2e^{-} -> Cu [/ce] [math]0.34V[/math]

 

The anode here would be the Copper, so you would reverse the reaction, and make the voltage negative.

 

The Question also gives the following information: "Calculate E for the cell assuming a pH of 7 and that all Cr and Cu containing ions have a concentration of 0.10M. Assume standard conditions"

 

We then apply the Nerst equation

[math] E = E^o - \frac{.0592}{n} logQ [/math]

 

[math]E^o[/math] would be 1.33V-.34V, to give .99V. And n is the moles of elecrons, 6, once the second equation is normalized by multiplying coefficents by 3.

 

However, Q is eluding me. I know it's the concentration of the anode/cathode, but I think the pH comes into play somewhere, and I'm not sure how. Thanks.

[Tartaglia]

Cr2O72- + 14H+ + 3Cu --> 2Cr3+ +7H2O + 3Cu2+

 

 

K = [Cr3+]^2*[H2O]^7*[Cu^2+]^3/{[Cr2O72-]*[H+]^14}

 

Since the solutions are reasonably dilute we can take the activities to be the concentrations

 

Cu is a solid and [Cu] is defined as 1, [H2O] is constant and [H+] is constant at 10^-7

 

Therefore K will change by a factor of 0.1^2*0.1^3/0.1 = 0.1^4

 

therefore ln Q = ln {0.1^4}

[woelen]

Tartaglia, are you sure about the concentration of H(+)? A solution of a dichromate is somewhat acidic. Besides that, the reaction consumes quite a lot of H(+), so soon, at the very low concentrations of H(+) involved here, the concentration of H(+) rapidly changes. A cell, based on this quickly changes properties (deteriorates, because the reduction of chromium (VI) to chromium (III) becomes much harder at increasing pH).

[Tartaglia]

Woelen - Good point the pH will change from standard conditions where [H+] = 1

 

therefore K will change by a factor of 0.1^2*0.1^3/{0.1*{10^-7}^14} = 10^94

 

lnQ = ln(10^94)

 

making emf about two volts less ie a lot less favourable

[Tartaglia]

the 0.0592 factor should be RT/F which I make 8.314*298/96500C = 0.02567

 

This would make the emf about 1V less - still substantial

[Tartaglia]

0.0592 = 0.02567 * ln 10 Thats why