Question:
A steady current of 1.00 ampere is passed through an electrolytic cell containing a 1 molar solution of AgNO3 and having a silver anode and a platinum cathode until 1.54 grams of silver is deposited.
(a)How long does the current flow to obtain this de¬posit?
(b)What weight of chromium would be deposited in a second cell containing 1–molar chromium(III) ni¬trate and having a chromium anode and a platinum cathode by the same current in the same time as was used in the silver cell?
One Response
(a) for silver nitrate to deposit Ag at cathode half reaction
look up standard electrode potentials table
Ag(+) + e —> Ag(s)
1 mole of electron is transferred per mole of Ag deposited
t = [ (m)(e)(F) ] / [ (I)(M) ]
t = time in sec
m = mass grams deposited 1.54g
e = 1 (1mole of electrons per mole Ag)
F = faraday constant 96485.3399 C/mol
I = current 1 amp
M = molar mass of Ag 107.87 g/mol
t = [ (1.54)(1)(96485.3399) ] / [ (1)(107.87) ]
t = 1377.5 sec
(b) for chromium nitrate solution half cell cathode reaction
look up standard electrode potentials table
Cr(3+) + 3e —> Cr(s)
e = 3
( 3 moles of electrons are transferred per mole of Cr deposited)
molar mass of Cr is 52 g/mol
t = [ (1.54)(3)(96485.3399) ] / [ (1)(52) ]
t = 8572.4 sec
x = (1.54)(1377.5) /(8572.4) = 0.247 gram