Question

(a) How many mole of mercury will be produced by electrolysing 1.0 MHg `(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 `Alk^(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 1

Time = 3 hours
`=3xx 60xx60= 10800 Sec`.
Current = 2A
Charge = Current `xx` time
`= 2 xx 10800 = 21600 C`
According to the questions
Solution of `Hg(NO_(3))_(2) = 1 mol`.
Given `Hg(NO_(3))_(2) = 200.g//mol`
We required 2F or `2 xx 96487C`to deposite 1 mole 63 g. of Hg
For 21600 C, the mass of mercury is
`=((200.6g//mol)xx (21600^(@)C))/((2xx 96487C//mol)) = (4332960 g)/(192975)=22.45g`
Moles `= (22.45)/(200.6)= 0.112 mol`
(b) Given
`E_(Ni^(2+)//Ni)^(2) = - 0.25V`, `E_(Al^(3+)//Al)^(o) = - 1.66V`
Half cell equations are
`Al to Al^(3+) + 3e^(-)" ""(at anode)"`
`Ni^(2) + 2e^(-) to Ni" ""(at Cathod)"`
`2Al + 3Ni^(2+) to 2Al^(3+) + 3Ni" ""over all recation"`
The cell may be represented as
`Al|Al^(3+)|| Ni^(2+) | Ni`
`E_("Cell")^(o) = E_("right")^(o) - E_("left")^(o)`
`=(-0.25)-(1.66)`
`= -0.25 + 1.66 rArr E_("cell")^(o) = 1.41 V`.