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5. Consider the galvanic cell, Pb(s) Pb(aq) || Cu(aq) Cu(s) What should be done to increase the cell potential (i.e., become
8. Balance the following reduction-oxidation reaction in an acidic medium. MnO,-(aq) + Cl-(aq) → Mn* (aq) + Cl2(g)
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Answer #1

#5:

The Nernst equation for the given Galvanic cell is:

Ecell = Eocell - (0.0591/n)*log([Pb2+(aq)] / [Cu2+(aq)])

In order to increase cell potential, Ecell, the value of log([Pb2+(aq)] / [Cu2+(aq)]) must decrease.

log([Pb2+(aq)] / [Cu2+(aq)]) will decrease when the ratio: [Pb2+(aq)] / [Cu2+(aq)] will decrease.

[Pb2+(aq)] / [Cu2+(aq)] will decrease when either [Cu2+(aq)] concentration is increased or [Pb2+(aq)] concentration is decreased

(Answer)

#6:

Since Pd2+(aq) has higher reduction potential, it undergoes reduction and Fe(s) undergoes oxidation. Hence

EoOxi = - (-0.45V) = 0.45 V

EoRed = 0.95 V

Eocell =  EoOxi + EoRed = 0.45V + 0.95V = 1.40 V

The Nernst equation for the given Galvanic cell is:

Ecell = Eocell - (0.0591/n)*log([Fe2+(aq)] / [Pd2+(aq)])

Since Pd2+(aq) is reduced to Pd(s), 2 electrons are transferred. Hence n = 2

=> Ecell = 1.40V - (0.0591/2)*log(0.100 / 1.0*10-5)

=>  Ecell = 1.2818 V (Answer)

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