Select correct option for the given statements (where T = true anf F= false ) i) electrolysis of dilute aqueous solution of NaCl is done with inert electrodes then pH of the solution if increases. ii) if electrolysis of dilute aqueous solution of NaCl is done with inert electrodes then on adding methyl orange to the solution it becomes orange coloured. iii) if electrolysis of concentrated aqueous solution of NaCl is done with inert electrodes pOH of the solution increases. iv) if electrolysis of concentrated aqueous solution of NaCl is done with inert electrodes then the solution obtained turns red litmus to blue
Electrochemistry MCQs for NEET — Chemistry Questions with Answers
Practice free Electrochemistry (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
If at $ 50 ^\circ C temperature value of E^0_{ox} is + 0.83 V for the half cell Pt ∣ H_{2(g, 1.0, bar)} | OH ^- _{(aq)} then what is the value of ionic product of water at 50 ^ \circ C $ ?
What quantity of electricity is required for complerte reduction of all $ Ag ^+ from 1.0 M AgNO_3 $ is 250 ml aqueous solution ?
The quantity of electricity required for the complete reduction of all Ag+ ions from 1.0 M AgNO3 solution of 250 ml is calculated using the formula: Q = nFz, where n is the number of moles of Ag+, F is the Faraday constant (96485 C/mol), and z is the charge on the Ag+ ion (1+). With the given concentration and volume, the number of moles of Ag+ is 0.25, so Q = 0.25 × 96485 × 1 = 24125 C.
Using 2 g Hg cathode Cd-Hg amalgam is obtained by electrolysis of $ CdCl_2$ then how much ampere electric current should be passed for 100 seconds to obtain Cd-Hg having 20% Cd ? (At.wt Cd = 112.5 g mol-1 )
To obtain a Cd-Hg amalgam with 20% Cd, the number of moles of Cd required is 0.2 × (2/112.5) = 0.00356 mol. Using the equation Q = nFz, where n is the number of moles (0.00356), F is the Faraday constant (96485 C/mol), and z is the charge on Cd2+ (2+), we get Q = 0.00356 × 96485 × 2 = 686 C. Since Q = It, where I is the current in amperes and t is time in seconds (100 s), we have I = Q/t = 686/100 = 8.58 A.
On passing 5 amp current for 2.15 hours through unknown solution of the salt of the Pd metal theoritically 10.64 g of Pd metal get deposited then what is the oxidation state of the Pd in that salt ( $ At.wt Pd = 106.4 g mol ^ {-1} $ )
If electrochemical equivalent of the metal is $ 4.0 \times 10 ^ {-4} g /coulomb $ then how much metal will be deposited at cathode on passing 15 amp current for 2 hours where current efficiency is 75 %
Theoritically how much quantity of electricity is required to obtained 112 ml of hydrogen gas at STP by electrolysis of acidic water ?
For how much times 0.5 A current should be passes through aqueous solution of CuSO_4 to deposit 2 g of Cu by electrolysis ? $ ( At.wt Cu = 63. 5 g mol ^ {-1}$ )
When aqueous solution of $ AgNO_3 $ is electrolysed with platinum electrodes its concentration decreases from 4M to 3 M. if same solution is electrolysed with Ag electrode then which of the following observation will be observed ?
If 0.5L 2.0 M aqueous solution of $ Ni(NO_3)_2 $ is electrolysed between graphite anode and nickel cathode by passing 9.65 A current for 3 hours then what will be the concentration of $ Ni(NO_3)_2 $ solution?
During electrolysis of Ni(NO₃)₂ solution with graphite anode and Ni cathode, Ni²⺠ions get discharged at the cathode, forming Ni metal. This decreases the concentration of Ni²⺠ions in the solution. The amount of Ni²⺠ions reduced can be calculated using the provided current and time data, leading to a final concentration of 0.92 M.
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