Home
Class 12
PHYSICS
A copper refining cell consists of two p...

A copper refining cell consists of two parallel copper plate electrodes 5 cm apart and 1 metre squre, immersed in a copper sulphate solution of resistivity `12xx10^(-2)Omegam`. Then the potential difference which must be established between the plates to provide a constant current to deposit `0.66kg` of copper on cathode in one hour is nearly `(Z=3.33xx10^(-7)kg//C)`

A

0.6V

B

0.33V

C

3.0V

D

15V

Text Solution

Verified by Experts

`m=Zit=Z((V)/(R))t`
`rARr V=(mR)/(Zt)=(mrhol)/(Zat)`
`=(0.66xx1.2xx10^(-2)xx5xx10^(-2))/(3.33xx10^(-7)xx1xx60xx60)`
=0.33 volt
Promotional Banner

Similar Questions

Explore conceptually related problems

A copper voltameter is connected in series with a coil of resistance 100Omega . A steady current flowing in the circuit for 10 minutes, gives a deposit of 0.1 g of copper in the voltameter. Calculate the heat generated in the resistance coil. (ECE of Cu 3.3xx10^(-7)kg//C )

The equivalent conductance of 0.10 N solution of MgCI_(2) is 97.1 mho cm^(2) eq^(-1) . A cell electrodes that are 1.50 cm^(2) in surface are and 0.50 cm apart is filled with 0.1N MgCI_(2) solution. How much current will flow when the potential difference between the electrodes is 5 volts?

There are 8.4xx10^(22) free electrons per cm^(3) in copper. The current in the wire is 0.21A (e= 1.6xx10^(-19)C ). Then the drifts velocity of electrns in a copper wire of 1mm^(2) cross section, will be:-

Two metal plates having a potential difference of 800 V are 0.02 m apart horizontally. A particle of mass 1.96 xx 10^(-15) kg is suspended in equilibrium between the plates. If e is the elementary charge, the charge on the particle is

Two metallic plate A and B , each of area 5 xx 10^(-4)m^(2) , are placed parallel to each at a separation of 1 cm plate B carries a positive charge of 33.7 xx 10^(-12) C A monocharonatic beam of light , with photoes of energy 5 eV each , starts falling on plate A at t = 0 so that 10^(16) photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 10^(6) incident photons fall on it per square meter per second. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remain constant at the value 2 eV Determine (a) the number of photoelectrons emitted up to i = 10s, (b) the magnitude of the electron field between the plate A and B at i = 10 s, and (c ) the kinetic energy of the most energotic photoelectrons emitted at i = 10 s whenit reaches plate B Negilect the time taken by the photoelectrons to reach plate B Take epsilon_(0) = 8.85 xx 10^(-12)C^(2)N- m^(2)

Suppose that the particle in Exercise in 1.33 is an electron projected with velocity v_x = 2.0 xx 10^6 ms^(-1) . If E between the plates separated by 0.5 cm is 9.1 xx 10^2 N/C, where will the electron strike the upper plate ? (|e| = 1.6 xx 10^(-19) C, m_(e) = 9.1 xx 10^(-31) kg)

The conductivity of pure water in a conductivity cell with electrodes of cross-sectional area 4cm^(2) placed at a distance 2cm apart is 8 xx 10^(-7) S cm^(-1) . Calculate. (a) The resistance of water. (b) The current that would flow through the cell under the applied potential difference of 1 volt.

Glycerine flows steadly through a horizontal tube of length 1.5 m and radius 1.0 cm. IF the amount of glycerine collected per second at one end is 4.0 times 10^-3 kg s^-1 . What is the pressure difference between the two ends of the tube?(density of glycerine =1.3 times 10^3 kg m^-3 and viscosity of glycerine =0.83 Pa s).[You may also like to check If the assumption of laminar flow in the tube is correct]

Glycerine flows steadily through a horizontal tube of length 1.5m and radius 1.0 cm . If the amount of glycerine collected per second at one is 4.0xx10^(-3)kgs^(-1) , what is the pressure difference between the two ends of the tube ? (Density of glycerine =1.3xx10^(10^(3)kgm^(-3) and viscosity of glycerine =0.83 Pa s). [ you may also like to check if the assumption of laminar flow in the tube is correct]

In the nuclear fusion reaction _(1)^(2)H + _(1)^(3)H rarr _(2)^(4)He + n given that the repulsive potential energy between the two nuclei is - 7.7 xx 10^(-14) J , the temperature at which the gases must be heated the reaction is nearly [Boltzmann's constant k = 1.38 xx 10^(-23) J//K]