# Consider a galvanic cell consisting of the following two redox couples:?

Ag+(0.010M) +e- -> Ag E= +0.80V

1.) Ag+ + e- –> Ag
The cathode is where reduction occurs. The lower the value of the standard reduction potential of a half-cell reaction, the more likely it is to be reduced. Because 0.20 V (Bi) is less than 0.80 V (Ag), Bi will be oxidized (As is seen in part 2)–And therefore, silver will be reduced.
2.) Bi –> Bi3+ + 3e-
3.) 3Ag+ + Bi –> Bi3+ + 3Ag (simply balance charge)
4.) At standard conditions, Ecell = Ecathode – Eanode
Ecell = 0.80 V – 0.20 V = 0.60 V
5.) E cathode = 0.80 V – 0.0592/3* log(1/[0.010]^3) = 0.682 V
E anode = 0.20 V – 0.0592/1*log(1/[0.0010]) = 0.0224 V
Ecell = 0.682 V – 0.0224 V = 0.704 V

1) remember: red cat (reduction occurs at the cathode) and an ox (oxidation occurs at the anode.)
So: the half rxn at cathode is: Ag+ + e- => Ag
2) see above. Bi=> Bi3+ + 3e-
The reason Bi gets oxidized and Ag gets reduced is because one has to get oxidized and one has to get reduced and only positive voltages will work in a galvanic cell, because only positive voltages are spontaneous. Therefore, you must reverse the 2nd half rxn and make the voltage negative; you get +.80-.20=.6V I hope this isn’t getting to hard to follow, more explaining in 4).
3) You pretty much just add the two revised half rxns together and balance. including charge.
So: 3Ag+ 3e-=> Bi3+ + 3Ag + 3e-
4) Since the equation given is the reduction rxn for Bi, and Bi is actually being oxidized in this cell, you just make the .20V negative and add it to the .8 of the Ag reduction. That gives a .60 V E cell. REMEMBER, coefficients DO NOT MATTER. I can’t stress this enough because tons of my friends have tripped up on that part.
5) oh crap, i hate these problems, you actually have to use the nernst equation:
E0cell-RT/nF(lnQ), in which: R=8.31, T=temperature in kelvin, n=moles of electrons transferred, and F=faraday’s constant, and E0cell is the standard cell potential.
.60-.[0592/(3)](logQ)=Ecell
so what is Q? Q is concentration of products to the power of the coefficient/concentrations of reactants to the power of the coefficient. So Q is just the Bi 3+ concentration (product) over the Ag concentration (reactant). Bi and Ag don’t matter because they are not ions or gases.
so: (.0010)/(.010)^3 I got 1000, but i don’t have a calculator with me right now so you better check that.
So you get: .60-.0592/(3) x (log1000)
since the log of 1000 is 3, the 3’s cancel out. .60-.0592= you figure that out. i’m no good at mental math.
unit would be volts, significant figures would be 2 (i think. I’m not good at those either)