posted by Joe on .
Determine the potential (in V) of an electrochemical cell using a gold electrode and gold (III) [0.0243 M] solution in combination with a manganese/manganese (II) [0.00567 M] half cell
How on earth would I do this question?
I just know
Au(3+) + 3e- = Au and E = 1.22 V
The half reactions are:
Mn^2+(aq) + 2e- --> Mn(s) Eo = -1.18
Au^3+(aq) + 3e- --> Au(s) Eo = 1.50***
(***NOTE: That is the value I was able to find)
Since Mn oxidizes more easily than Au, the Mn/Mn^2+ half reaction occurs as an oxidation:
Mn(s) --> Mn^2+(aq) + 2e- Eo = +1.18
The overall reaction is:
2Au^3+(aq) + 3Mn(s) --> 2Au(s) + 3Mn^2+(aq)
Keq = [Mn+2]^3/[Au+3]^2
The Eo (standard potential for the entire electrochemical cell is 1.50 + 1.18 = 2.68 volts
Eo(overall) = 2.68v
The standard cell potential assumes standard conditions: 1M solutions and 1 atm pressure for gasses present.
Since you DO NOT HAVE 1M solutions, you must use the Nernst Equation:
E(cell) = Eo - [0.0592/n][log(Q)]
Q = [Mn+2]^3 / [Al+3]^2
(Same as the Keq BUT the concentrations are the initial ones rather than the equilibrium ones)
E(cell) = 2.68 - [0.0592/6][log[(0.00567)^3/(0.0243^2)] = _______?
Awesome Thanks So Much.
But.. how do you know that in the Nernst Equation n = 6
Besides that it makes sense.. thanks again!
The Mn rxn is 2e, the Au rxn is 3e.
Multiply Mn rxn by 3 and Au rxn by 2 and add the two equations together. That gives 6 electrons; therefore, n is 6 in the reaction.
The two half reaction must be reconciled on the number of electrons by multiplying the first one by 3 and the second one by 2:
3Mn(s) --> 3Mn^2+(aq) + 6e-
2Au^3+(aq) + 6e- --> 2Au(s)
3Mn(s) + 2Au^3+(aq) + 6e- --> 3Mn^2+(aq) + 2Au(s) + 6e-
The 6e-'s on each side are usually cancelled out in the overall reaction but they can't be forgotten.
Wow you guys are awesome. Thanks!!