18. NEED HELP FAST

Question

Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that  [Zn2+]=0.00100 M  and  [Au3+]=0.871 M . Standard reduction potentials can be found in this table.

3Zn(s)+2Au3+(aq)↽−−⇀3Zn2+(aq)+2Au(s)

𝐸=

Answer 1

For this problem, we need to use the Nernst Equation.

Ecell = E° cell - (RT)/(nF)*ln([red]/[ox]), where R is the universal gas constant, T is temperature in kelvin, n is the number of electrons transferred in the reaction, F is Faraday's constant, [red] is the concentration of the ion at the cathode, and [ox] is the concentration of the ion at the anode.

We can calculate E° cell from the table of standard reduction potentials.

E° cell = E° red, cathode - E° red, anode = 1.5 - (-.76) = 2.26 V

The standard reduction potential at the cathode corresponds to the reduction half reaction Au³⁺ + 3e- → Au

The standard reduction potential at the anode corresponds to the reduction half reaction Zn²⁺ + 2e- → Zn

Notice the E° red, anode is the reverse of the oxidation half reaction in the given chemical equation.

We can then plug this value into the Nernst equation to solve for the cell potential. Use 6 for n because the reduction half reaction, involving 3 electrons must be multiplied by 2 and the oxidation half reaction, involving 2 electrons must be multiplied by 3 so the total electrons, 6, cancel out.

Ecell = 2.26 - (8.31*298)/(6*96485)*ln(.871/.001)= 2.23. V