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physics problems

physics problems

by arvind ramachandran -
Number of replies: 1

A steady E.M.F of 1.62V is maintained acresstwo platinum electrodes placed in a solution of CuCl2.At the end of 600 sec, the mass of Cu deposited is 3.175g.The back E.M.F of voltameter is given to be 1.427V.Calculate the resistance of the voltameter.

The Potential Diff. across the terminals of a bATTERY OF e.m.f 12V
and int.resistance 2 ohm drops to 10V when it is connected to Ag Voltameter.Cal the silver deposited at the cathode in half an hour.
(At.wt of Ag is 107.9)

A Ag and Cu voltameter are connected in series with a 12V battery of negligible int.R,0.806g of Ag is deposited in half an hour in the Ag voltameter.Calculate
i)magnitude of current flowing in the circuit
ii)mass of Cu deposited in the Cu voltameter during the same
[ECE of Ag=1.12x10^-8 kg/C
 ECE of Cu=6.6x10^-7  kg/C]

In reply to arvind ramachandran

Re: physics problems

by Asif Zafar -

1 Faraday is the charge required to displace a mole of electrons.

1 F = 96500C where C stands for Coulumbs .

Let (MM) stand for Molecular Mass or Atomic Weight.

(i) stands for current, (t) stands for time and (q) for charge.

  1. The resistance (R) of voltmeter is equal to:                                 (back emf of voltmeter)/(currents passing through the voltmeter). In this case q=(3.175)*2*F/( MM of Cu ) [*2 coz' Cu deposists in Cu+2 state]. Now u can calculate  i=q/t. Therefore R=1.427*t/ q.
  2. Since internal R of 2 ohms produces a drop of 2 volts the current flowing through the circuit is 1 amp. Therefore q accumulated in half an hour is q=i*t=1*1800=1800. Therefore :                                        wt. of Ag deposited=1800*(At. Wt. of Ag)/F
  3. i=0.806*F/[(At. Wt. of Ag*t].                                                        (Wt. of Cu deposited)=i*t*(At. Wt. of Cu )/ [2*F].      smile