# A current of 0.02A is passed for 2.5hrs through a solution of a salt of the metal M. The mass of M deposited at the cathode is 0.0349. The relative atomic mass of M being 56, what is the charge on the ions of M? This electrolysis problem will start with the determination of how many moles of electrons are used in the experiment.

Important details:

1A = C/s

1F = 96485 C = 1 mole of electrons

1. We get the total C used

(0.02 C/seconds) (2.5 hours x 60min/hours x 60 seconds/minutes)

= 180 C

2. get the mole of electrons

180 C x ( 1F / 96485 C) x (1mole e/1F )

= 1.86557x10^-3 mole electrons

3. Get the moles of the metal M used in the electrolysis

moles = mass compound/molar mass

= 0.0349 grams/ 56 grams/mole

= 6.2321 x10^-4 moles metal M

4. There are 1.86557x10^-3 mole electrons that were transferred during the electrolysis and 6.2321 x10^-4 moles metal M used. To get the charge of the metal we have to get the number of electrons that have transferred.

# of electrons = moles electron/ moles metal M

= 1.86557x10^-3 / 6.2321 x10^-4

= 2.99 = 3 electrons

Therefore there are 3 electrons that have transferred or we can say that the metal M loses 3 electrons. Thus the charge of the metal M is 3+. We can write the equation as:

M ---> M^3+ + 3e'

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