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.
Oh Okay!
Wow you guys are awesome. Thanks!!
To determine the potential of the electrochemical cell, you can use the Nernst equation, which relates the potential of a cell to the concentrations of the reactants. The equation is as follows:
E_cell = E_reduced - E_oxidized
= E_cathode - E_anode
= E_cathode - E_metal
In this case, the gold electrode is serving as the cathode and the manganese/manganese (II) half-cell is serving as the anode.
Given:
E(cathode) = 1.22 V (the reduction potential of Au(3+) to Au)
[Au(3+)] = 0.0243 M (concentration of gold(III) solution)
[Mn(II)] = 0.00567 M (concentration of manganese(II) solution)
To complete the calculations, you need the reduction potential of the manganese/manganese (II) half-cell, which is not provided in your question. The reduction potentials of half-cells are typically provided in tables, so you can look up the value for Mn(II).
Once you have the reduction potential of Mn(II), you can use the Nernst equation to calculate the potential (in V) of the electrochemical cell as follows:
E_cell = E_cathode - E_anode
= (E_reduction of Au(3+) to Au) - (E_reduction of Mn(II)...)
Using the Nernst equation:
E_cell = E_cell - (0.05916 V/n) * log([Au(3+)] / [Mn(II)])
= 1.22 V - (0.05916 V/3) * log(0.0243 M / [Mn(II)]) [since the balanced equation for Au(3+) reduction involves 3 electrons]
By substituting the known values and the value you find for the reduction potential of Mn(II), you can calculate the potential of the electrochemical cell in volts (V).