Question

A galvanic cell has the following composition:

Pt(s) l Sn4+ (1 M), Sn2+ (1 M) ll Cu2+ (1 M) l Cu(s)

From the shorthand notation, draw the galvanic cell. Label the anode and cathode and indicate which species are in the aqueous phase and which are solids (the electrodes). Under each beaker, write the appropriate half-rxn.

Answer 1

Reduction = species that GAINS electrons

Oxidation = process in which a specie will LOSS electrons

Reducing agent = The species that favors reduction, i.e. it will oxidize in order to reduce another species

Oxidizing agent = The species that favors oxidation, i.e. it will reduce in order to oxidise another species

The Galvanic Cell favors this red-ox process via:

The anode = electrode in which the OXIDATION takes place, therefore, the reducing agent will get oxidized (lose electrons). The electrons flow from the solid electrode to the voltmeter flow. The solid species will then turn ionic, and go into solution. Since the solid converts to ion in solution, then this electrode will lose mass as time passes by.

The cathode = electrode in which REDUCTION takes place, therefore, the oxidizing agent will get reduced (gain electrons). The electrons are received from the voltmeter ( which come from the anode ). The ions in solution will then form solid over the cathode. Since there is solid formation, there will be an increase in the mass of this electrode.

Electrons, therefore, flow from oxidation cell to reduction cell; that is, from the anode to cathode.

Note that, as expected, there will be a charge overload if we kept this cell running as it is. Therefore, we must add a salt bridge, which allows ionic exchange, i.e. the charges will balance each other, positive ions and negative ions will flow in order to balance the cell’s charge.

then

anode = Sn4+/Sn2+ set

cathode = Cu(s)

solid electrode is copper

aqueous electrode is Sn

half reaction(oxidation/anode)

Sn2+ = Sn4+ + 2e-

half reaction(reduction/cathode)

Cu+2 + 2e- = Cu(s)