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?
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This electrolysis problem will start with the determination of how many moles of electrons are used in the experiment.
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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