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Question

Which of the following of metals, when coupled, will give maximum EMF for a voltaic cell?
Given:
Au+(aq)+eAu(s); E0=1.69 V
Cu2+(aq)+2eCu(s); E0=0.34 V
Pb2+(aq)+2ePb(s); E0=0.13 V
Fe2+(aq)+2eFe(s); E0=0.44 V
Ca2+(aq)+2eCa(s); E0=2.87 V

A
Fe and Au
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B
Pb and Au
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C
Cu and Au
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D
Ca and Cu
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Solution

The correct option is D Ca and Cu
Standard reduction potentials (SRP) of the half cell electrodes are calculated and arranged in a series of increasing order which is known as electrochemical series.
The standard electrode potential of H+/H2 couple is zero.

Here, a negative E0 signifies that it has high tendency to get oxidised and the redox couple is a stronger reducing agent than the H+/H2 couple.

A positive E0 signifies that it has high tendency to get reduced and the redox couple is a stronger oxidising agent than the H+/H2 couple.

Electrochemical series helps in the selection of electrode assemblies to construct the galvanic cells of desired EMFs.

For a galvanic cell to provide high cell voltage, the half cell having higher reduction potential and lower reduction potential should be coupled.

In given electrochemical series,
Au+(aq)+eAu(s) (Highest value)
Ca2+(aq)+2eCa(s) (Lowest value)

E0cell=SRP of substance reducedSRP of substance oxidised
(SRP - Standard reduction potential)

E0cell=E0cathode (red)E0anode (red)

For (a),
E0cell=1.69(0.44)
E0cell=2.13 V

For (b),
E0cell=1.69(0.13)
E0cell=1.82 V

For (c),
E0cell=1.690.34
E0cell=1.35 V

For (d),
E0cell=0.34(2.87)
E0cell=3.21 V

Thus, the SRP difference will be maximum for Ca and Cu.
Hence, it gives the maximum EMF for a voltaic cell.

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