InterviewSolution
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InterviewSolution
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(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`) |
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Answer» 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`. |
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