Question
Question: How much time will it take for a uniform current of 6.00A to deposit 78g of gold from a solution of ...
How much time will it take for a uniform current of 6.00A to deposit 78g of gold from a solution of AuCl4−.? What mass of chlorine gas will be formed simultaneously at the anode of the cell? (Atomic mass of Au =197)
Solution
Find the moles of Au needed to deposit 78g of Gold. Now multiply it by a change in oxidation number change in the reaction. The amount of charge to be transferred will be (moles of A) x (change in oxidation number). Divide it by the value of current and you will have the time. Write the reaction and see what amount of Cl2 released for a given amount of Au deposited
Complete step by step solution:
Calculation of oxidation number of Au in AuCl4−
Let oxidation number of Au be x
⟹ x -4(For each Cl) = -1
⟹ x = +3
⟹ The oxidation number of Au is +3
When Au is deposited it gets reduced to 0, so change in oxidation number of Au is 3
⟹ the number of electrons needed to reduce(deposit) one atom of Au is 3
Moles of Au required to deposit 78g of Au = Mm
Where m is mass of Au to be deposited
M is molar mass of Au
⟹n n = 19778
The amount of current required = n x 3 x NA x e = I t
NA is the Avogadro number
e is charge on electron
Also NA x e = F
F is Faraday’s Constant whose value is approximately equal to 96485 C/mol
I is the current, t is time
Putting all known values gives t=197×63×78×96485
⟹ t= 19101 sec.
Cl−1 in the given compound will reduce to Cl2
One mole of 2F will give one mole of Cl2
⟹ mass of Cl2 liberated = 2n×3×71 2n×3×71
71g is Molar mass of Cl2
. ⟹ Mass of chlorine gas liberated = 42.16 g
So, for the given options option D is the closest hence it is the answer.
Note: All of the Cl from the given compound will not be liberated as the number of electrons required for all 4 chlorine atoms to reduce is not sufficient. Also, note that the above formula used is the mathematical representation of Faraday’s first law of electrolysis.