[arvind1990]

Can anyone help me with the following problems

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]

[Donaldson Tan]

I love to help, but I don't know how to take in account of back EMF.

but I know this equation should help:

E (energy) = Q x V (charge x voltage)

where Q = I x t (current x time)

I am not a physicist or an electrical engineer.

However, I think V = EMF_cell - EMF_back