Answer:
Option A
Explanation:
Relationship obtained on combining Faraday's first and second laws.
$W=ZQ=\frac{E}{F}\times Q= \frac{Q}{F}\times E= \frac{Q}{F}\times \frac{N}{z}$
$=\frac{l\times t}{F}\times \frac{M}{z}$
Here, Z= electrochemical equivalent
Q= quantity of electricity passed
M= atomic mass
z= valency of the metal
F= one Faraday or 96500 C
Given, weight of calcium that has to be deposited=10 g
z of the metal ion=2
Molecular mass of calcium =40 g mol-1
$W=\frac{l. T}{F}\times \frac{M}{z}; 10=\frac{l. T}{F}\times \frac{40}{2}$
$=\frac{10\times2}{40}F=IT\Rightarrow 0.5F=l.T$
