[jaysup_2006]

The SCE is composed of mercury in contact with a saturated solution of calomel (Hg2Cl2). The electrolyte solution is saturated KCl. SCE is +0.242 V relative to the standard hydrogen electrode. Calculate the potential for each of the following galvanic cells containing a saturated calomel electrode and the given half-cell components at standard conditions.

For Cu2+ + 2 e- --> Cu

So i understand the voltage of a saturated cell is .242. The Voltage for the above reaction is .34. Would I add the two voltage, since the Cu2+ is saturated or would I use the nest equation(however no concentration is given, thats why i dont think so). I get confused here. Any ideas?

[Borek]

the given half-cell components at standard conditions.

It means activities of all forms equal 1.

[jaysup_2006]

so even combined with the SCE, the potenial is always going to be 1? I thought the .34V is the value for Cu2+ + 2 e- --> Cu at standard temp is the half reaction value(book given). So i wouldnt account for that value at all? Or would I subtract from the 1?

[Borek]

No. Activity (call it concentration if you like, they are related but not identical) is a number that't you will plug into Nernst equation:

E = E₀ + RT/(nF) ln Q

where Q is reaction quotient.

In SCE concentrations differ from 1 - that's the way electrode is built, but the potential you are given already takes care of that.

Cu/Cu²⁺ system potential depends on the Cu²⁺ concentration. Standard conditions mean [Cu²⁺] = 1, thus when plugged into Nernst equation expression under log is 1, and log value is 0.

[jaysup_2006]

Ok, so my equation would be set up like this then:

E = E0 + RT/(nF) ln Q =; E = E0 - .0591/n log(q); = E = .34 - 0591/2 log(1); E = .34-0

So .34V would be my final answer?

[Borek]

Close - but this only for Cu/Cu²⁺ half cell. Now you have to take SCE potential into account. Note that SCE potential is JUST given - it already contains correct value of Q (that's thanks to the way SCE is made).

[jaysup_2006]

ok...so since SCE has the lower voltage it gets run in reverse....so .34V - .242V = .098 is the new voltage