(may be broke/outdated!)
In an electrolytic cell using inert electrodes, a current of 4.00 A is passed through a molten binary salt with the formula XCl for 20.0 minutes.
1.15 g of metal X is produced at the cathode.
What is the metal X?
(a) Li
(b)Na
(c)K
(d)Rb
(e)Cs
I tried to use ne=IT/F with F=9.65×10^4 c/mol, I=4c/s, T=20min=1200s.
Then divided 1.15 g with calculated ne, 4.97×10^-2 mol.
The answer I got was about 23.14 g/mol, and closest was Na with 22.99 g/mol. Did I get correct or did something wrong??
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3 Responses
You are all good (.15 off?). I’d say Na is the correct answer
M+ + e- —> M, one faraday (a charge of one mole electrons) that is 96,500 coulombs produce one mole of the metal. Q= 4 x 20 x60 = 4800 coulombs. 4800 coulombs produce 1.15 grams metal, 96,500 coulombs current produce 23.11 grams of metal. It is the atomic weight of the metal . Hence the metal is sodium. You got correct .
Q=It = 4 x 20 x 60 = 4800 C
Q= n(e-)xF so n(e-) = Q/F = 4800/96485 = 0.049749 mol
As X is a group 1A alkali metal (because Cl- needs X+ to be balanced) then the formula for the reduction of the metal (from +1 oxidation to 0 oxidation is:
X+(aq) + e- ——–> X (s)
So 1 mol of e- for 1 mol of X (s) formed so we can calculate the formula weight:
FW = m/n = 1.15g / 0.049749 =23.11 g/mol
So as you said, the closest answer is Na.