Question

The following question is a case-based question. Read the case carefully and answer the questions that follow.

In a galvanic cell, chemical energy of a redox reaction is converted into electrical energy, whereas in an electrolytic cell the redox reaction occurs on passing electricity. The simplest galvanic cell is in which \[\ce{Zn}\] rod is placed in a solution of \[\ce{ZnSO4}\] and \[\ce{Cu}\] rod is placed in a solution of \[\ce{CuSO4}\]. The two rods are connected by a metallic wire through a voltmeter. The two solutions are joined by a salt bridge. The difference between the two electrode potentials of the two electrodes is known as electromotive force. In the process of electrolysis, the decomposition of a substance takes place by passing an electric current. One mole of electric charge when passed through a cell will discharge half a mole of a divalent metal ion such as \[\ce{Cu^2+}\]. This was first formulated by Faraday in the form of laws of electrolysis.

Answer the following questions:

(a) What is the function of a salt bridge in a galvanic cell?   (1)

(b) When does galvanic cell behave like an electrolytic cell?  (1)

(c) Can copper sulphate solution be stored in a pot made of zinc? Explain with the help of the value of \[\ce{E^0_{cell}}\].  (2)

\[\ce{(E^0 Cu^2+/Cu = 0.34V)}\]

\[\ce{(E^0 Zn^2+/Zn = -0.76V)}\]

                                                 OR

(c) How much charge in terms of Faraday is required for the following:  (2)

1. 1 mol of \[\ce{MnO^-4}\] to \[\ce{Mn^2+}\]
2. 1 mol of \[\ce{H2O}\] to \[\ce{O2}\]

Answer 1

(a) It maintains electrical neutrality and completes the circuit.

(b) When the external potential exceeds the reaction's voltage, the reaction begins to proceed in the opposite direction. At this stage, the cell transforms into an electrolytic cell, which employs electrical energy to perform a chemical process that does not occur spontaneously. 

(c) No, a spontaneous substitution reaction will occur between copper sulphate and zinc in a zinc pot, with zinc dislodging copper to generate zinc sulphate. 

\[\ce{CuSO4 + Zn -> Cu + ZnSO4}\]

Zinc functions as an anode because its reduction potential is lower than copper's. Zinc will lose two electrons as a result of the oxidation at the anode with copper as the cathode, which has a high potential for reduction, will either accept electrons or undergo reduction.

\[\ce{E^0_{cell} = E^0_{cathode} - E^0_{anode}}\]

= 0.34 − (−0.76)

= 1.10 V

                                   OR

(c) i. \[\ce{MnO^-4 (+7) -> Mn^2+}\] 

Faraday's number required = change in oxidation number

= 7 − 2

= 5 F   ...[∵ F = 965000]

ii. \[\ce{H2O -> O2}\]

Change in oxidation number is 2.

So, No. of faraday's required is 2 or 2 F.