Saved Bookmarks
| 1. |
(a) How many mole of mercury will be produced by electrolysing 1.0 M Hg `(NO_(3))_(2)` solution with a current of 2.00 A for 3 hours? [Hg`(NO_(3))_(2) = 200.6 g mol^(-1)`]. (b) A voltaic cell is set up at `25^(@)`C with the following half-cells `Al^(3+)` (0.001M) and `Ni^(2+)` (0.50M). Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential. (Given : `E_(Ni^(2+)//Ni)^(2)=- 0.25 V,E_(Al^(3+)//Al)^(@)= - 1.66V`) |
|
Answer» (a). `m=ZxxIxxt` `m=(200.6xx2xx3xx60xx60)/(2xx96500)` `m=(43329.6)/(1930)=22.45g` Number of moles `=(22.45)/(200.6)=0.112` mole (b). `[Al(s)toAl^(3+)(aq)+3e^(-)]xx2` `underline([Ni^(2+)(aq)+2e^(-)toNi(s)]xx3)` `2Al(s)+3Ni^(2+)(aq)to2Al^(3+)(aq)+3Ni(s)` `E_(cell)=(E_(Ni^(2+)//Ni)^(0)-E_(Al^(3+)//Al)^(0))-(0.0591)/(6)log(([Al^(3+)]^(2))/([Ni^(2+)]^(3)))` `=[-0.25V-(-1.66V)-(0.0591)/(6)log((10^(-3))^(2))/((0.50)^(3)))` `=1.41V-(0.0591)/(6)log((8xx10^(-6))/(1)` `=1.41-(0.0591)/(6)log[8+log10^(-6)]` `=1.41V-(0.0591)/(6)xx-5.0969` `=1.41V-(0.0591)/(6)` `=1.41V+0.0502V=1.4602V` |
|