A steady current of 1.00 ampere is passed through an electrolytic cell containing a?

Question

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) nitrate and having a chromium anode and a platinum cathode by the same current in the same time as was used in the silver cell?

(c) If both electrodes were platinum in this second cell, what volume of O2 gas measured at standard tempera¬ture and pressure would be released at the anode while the chromium is being deposited at the cathode? The current and the time are the same as in (b)

Answer 1

n(Ag)=0.0143 mol

since Ag oxidation number is 2+, it need twice the amount of mols of electrons to precipitate

so n(e)=0.0286 mol

(a) equations to use Q=i*t & Q=n(e)*F

>which can be written as: i*t=n(e)*F

>rearranging that equation and get t=n*F/i

>sub in values t=0.0286*96500/1

>t=2760 seconds (rounded)

(b) t=2759.9 s, i = 1 A, n(e) = ?

>n(e) = i*t/F = 1*2759.9/96500 = 0.0286 mol

>n(e):n(Cr)=3:1

>n(Cr)=0.009533 mol

>0.009533*52=0.496 grams

(c) n(e)=0.0286 mol

>n(e):n(O2)=4:1 since one O lose 2e

>n(O2)=0.00714 mol

>PV=nRT

**assuming standard temperature is 20oC, 101.325 kPa

>V=nRT/P

>V=0.00714*8.31*283/101.325=0.166 L