Question

The maximum current can be drawn from which of the following cells? ​

Answer 1

The given cells deals in Zinc and Copper half cells. Herein, Zn is being converted to Zn2+ and Cu2+ being converted to Cu.

Thus, the conversion equations are:

Zn - 2e --------» Zn2+

Cu2+ + 2e ----» Cu

=> z = 2

EMF of such a cell is:

E = E° - 0.059(V/z)[log(Co/Cr)]

E = E° - 0.059/2 ×[log(Zn2+/Cu2+)]

You see, the final EMF of the cell (E) totally depends upon the factor "-log(Zn2+/Cu2+)". So, for E to be maximum "log(Zn2+/Cu2+)" must be minimum.

See, the logarithmic function is an increasing function throughout it's domain. Thus, for "log (Zn2+/Cu2+)" to be minimum, the factor "Zn2+/Cu2+" must be minimum.

Now, let's calculate the magnitudes for "Zn2+/Cu2+" for the given options:

1 ) Zn2+/Cu2+ = 0.2/0.2 = 1

2) Zn2+/Cu2+ = 0.002/0.2 = 0.01

3) Zn2+/Cu2+ = 0.2/0.002 = 100

Clearly, from the above three calculated values of Zn2+/Cu2+ , the magnitude for second option is lowest.

Thus, the correct answer is Option(2).

Answer 2

Answer:

The given cells deals in Zinc and Copper half cells. Herein, Zn is being converted to Zn2+ and Cu2+ being converted to Cu.

Thus, the conversion equations are:

Zn - 2e --------» Zn2+

Cu2+ + 2e ----» Cu

=> z = 2

EMF of such a cell is:

E = E° - 0.059(V/z)[log(Co/Cr)]

E = E° - 0.059/2 ×[log(Zn2+/Cu2+)]

You see, the final EMF of the cell (E) totally depends upon the factor "-log(Zn2+/Cu2+)". So, for E to be maximum "log(Zn2+/Cu2+)" must be minimum.

See, the logarithmic function is an increasing function throughout it's domain. Thus, for "log (Zn2+/Cu2+)" to be minimum, the factor "Zn2+/Cu2+" must be minimum.

Now, let's calculate the magnitudes for "Zn2+/Cu2+" for the given options:

1 ) Zn2+/Cu2+ = 0.2/0.2 = 1

2) Zn2+/Cu2+ = 0.002/0.2 = 0.01

3) Zn2+/Cu2+ = 0.2/0.002 = 100

Clearly, from the above three calculated values of Zn2+/Cu2+ , the magnitude for second option is lowest.

Thus, the correct answer is Option(2).