Part A) A voltaic cell is set-up with a silver electrode in a silver nitrate sol

Question

Part A) A voltaic cell is set-up with a silver electrode in a silver nitrate solution and a nickel electrode in a nickel (II) nitrate solution. What is the reaction occuring at the anode?   Use the reduction potentials table in the notes/book.

Ni(s) Ni2+(aq) + 2e-

Ag+(aq) + e- Ag(s)

Ag(s) Ag+(aq) + e-

Ni2+(aq) + 2e- Ni(s)

Part B)What is the standard change in gibbs free energy for the following reaction in kJ?

2NO3-(aq) + 8H+(aq) + 3Cu(s) 3Cu2+(aq) + 2NO(g) + 4H2O(l)

Ni(s) Ni2+(aq) + 2e-

Ag+(aq) + e- Ag(s)

Ag(s) Ag+(aq) + e-

Ni2+(aq) + 2e- Ni(s)

Explanation / Answer

part A

2Ag + Ni2+ --> 2Ag+ + Ni this is the reaction mens silver under go oxidation and Ni under go reduction at anode always oxidation will takes place

so Ag(s) Ag+(aq) + e-

first we have to find out the standerd reduction potential of the cell

The two basic half-reactions (found in a table of standard electrode potentials) are:

Cu ==> Cu2+ + 2e- . . . . . . . . . . . . . .Eo = -0.34 V (cathode)
NO3- + 4H+ + 3e- ==> NO + 2H2O . . Eo = +0.96 V (anode)

Eo cell = E cathode + E anode = -0.34 V + 0.96 V = +0.62 V

now we have to use the following equation for calculatig the what is the standard change in gibbs free energy for the following reaction

Go=nFEo

Notice that this reaction involves a 6-electron change (i.e. Cu ==> Cu2+ + 2e- is a 2-electron change, so 3 of those (3Cu ==> 3Cu2+) is a 6-electron change).

Go=6x96485x0.62

Go= -358924.2 joules

Go= -358924.2/1000 kJ = -358.92kJ

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