Question

If \[\ce{E^{\circ}_{{Ag^{+}/{Ag}}}}\] = +0.80 V and \[\ce{E^{\circ}_{{Sn^{2+}/{Sn}}}}\] = −0.14 V, the standard emf of the cell \[\ce{Sn | Sn^2+ ||  Ag+ | Ag}\], will be ______.

Options

• 0.66 V

• 0.80 V

• 0.94 V

• 1.80 V

Answer 1

If \[\ce{E^{\circ}_{{Ag^{+}/{Ag}}}}\] = +0.80 V and \[\ce{E^{\circ}_{{Sn^{2+}/{Sn}}}}\] = −0.14 V, the standard emf of the cell \[\ce{Sn | Sn^2+ ||  Ag+ | Ag}\], will be 0.94 V.

Explanation:

The decrease potential is being considered. According to the electrochemical series, the higher the reduction potential, the greater the capacity of the metal in the electrode to be reduced. That is, it will function as a cathode. As stated in the question, Ag has a higher standard reduction potential; hence, it will act as a cathode, whereas Sn would act as an anode.

The EMF of the cell can be written as:

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

= 0.80 − (−0.14)

= 0.94 V.