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Question

Three ampere current was passed through an aqueous solution of an unknown salt of metal M for 1 hour. 2.997 gm of Mn+ was deposited at cathode. The value of n is (Atomic weight of M = 106.4)
  1. 1
  2. 4
  3. 2
  4. 3


Solution

The correct option is B 4
First always write down the half-reaction
Mn++neMRecall that W=I×t×E96500

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=106.4n
Plugging all this in, we have
2.997106.4n=3×1×60×6096500

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

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