Question

40. The maximum current can be drawn from which of
the following cells?
(1) Zn(s) /Zn2+(0.2M) //Cu2+(0.2M)/ Cu(s)
(2) Zn(s) /Zn2+(0.002M) //Cu2+ (0.2M) Cu(s)
(3)Zn(s)/Zn2+(0.2M)//Cu2+(0.002M)/Cu(s)
(4) All of these
​

Answer 1

Explanation:

Just calculate the emf of the cell by using nernst equation.

$e = 1.11 - \frac{0.059}{2} log( \frac{0.002}{0.2} ) \\ e = 1.1 - \frac{0.059}{2} log( {10}^{ - 2 } ) \\ e = 1.1 + 0.059 \\ e = 1.159volts$

Hence maximum emf will be in (2) optionand maximum current will be in case of (2)