1. To make a voltaic cell, a strip of copper is placed in a 1 M solution of copp

Question

1. To make a voltaic cell, a strip of copper is placed in a 1 M solution of copper nitrate and a strip of silver is placed in a 1 M solution of silver nitrate. The two metal strips are connected to a voltmeter by wires and a salt bridge is used to connect the solutions. These standard reduction potentials are found in a table:

When everything is connected, which of the following does NOT take place?

The copper half cell is the anode.

b. Copper metal is being produced in the copper half cell.

c. The concentration of silver ions is decreasing in the silver half cell.

d. Anions and cations move through the salt bridge.

2. Using the table of half-cell potentials from your textbook, which of these species is the most easily oxidized (is the best reducing agent)?

Pb, Ag, Sn2+, F-

The copper half cell is the anode.

Explanation / Answer

1) For the given Voltaic cell, "Cu" half cell is an anode while Ag half cell acts as cathode. Hence for oxidation half cell "Cu" electrod converts to Cu2+. Thus option b) will not occur in this case.

2) From standard reduction potentials -

E0 (Pb2+/Pb) = -0.126 V

E0(Ag+/Ag) = 0.80 V

E0(Sn+4/Sn+2) = 0.154 V

E0(F2/F-) = 2.87 V

These are all reducing agents, so the element which has less reduction potential value can easily get oxidize.

3) Since -

DG0 = - nFE0, DG0 = -RTlnKq

Therefore, nFE0 = RTlnKeq

                ln Keq = nFE0 / RT = 6*96500*0.122 / 8.314*298

                Keq = 2.41*1012

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