Question

Three ampere current was passed through an aqueous solution of an unknown salt of metal M for 1 hour. 2.997 gm of $M^{n+}$ was deposited at cathode. The value of n is (Atomic weight of M = 106.4)

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

First always write down the half-reaction
$M^{n+}+n{e}^{-}\to MRecallthatW=I\times t\times \frac{E}{96500}$

Here we are given that t = 1 hour and I = 3 A and we are also told that 2.997 grams of the ion was deposited.
So, W = 2.997 gm

But what is E? E is the molar mass/ n, therefore $E=\frac{106.4}{n}$
Plugging all this in, we have
$\frac{\frac{2.997}{106.4}}{n}=3\times 1\times 60\times \frac{60}{96500}$

Remember that here we need to input time in seconds and not hours
So, simplifying we will have n = 4