(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`)

Explanation

(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`