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Find the charge in coulombs required to convert 0.2 mole \(VO_{3}^{-2}\) into \(VO_{4}^{-3}\) –
Charge = 0.2 × 1 Faraday = 0.2× 96500 coulombs =19300 = 1.93 × 104 coulombs
When 9650 coulombs of electricity is passed through a solution of copper sulphate, the amount of copper deposited is (given at. wt. of Cu = 63.6)
Which of the following will form a cell with the highest voltage?
An electrolytic cell contains a solution of Ag2SO4 and has platinum electrodes. A current is passed until 1.6 gm of O2 has been liberated at anode. The amount of silver deposited at cathode would be
When electric current is passed through acidified water,112 ml of hydrogen gas at STP collected at the cathode in 965 seconds. The current passed in amperes is
Aluminium oxide may be electrolysed at 1000°C to furnish aluminium metal (At. Mass =27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is– Al3+ + 3e– ⟶ Al. To prepare 5.12 kg of aluminium metal by this method we require electricity of
Using same quantity of current, which among Na, Mg, and Al is deposited to the highest extent during electrolysis.
According to Faraday's law of electrolysis, the amount of metal deposited on an electrode is directly proportional to the amount of electricity passed and equivalent weight of metal. Now equivalent weight of sodium is 23, magnesium is \(\frac{24.3}{2}\) i.e. 12.15 and of aluminium is \(\frac{27}{3}\) i.e. 9. So, sodium which has highest equivalent weight among Na,Mg and Al, is deposited to highest extent during electrolysis.
If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode,is (eq.wt.of silver nitrate = 108)
b
A 0.5 M NaOH solution offers a resistance of 31.6 ohm in a conductivity cell at room temperature. What shall be the approximate molar conductance of this NaOH solution if cell constant of the cell is 0.367 cm-1.