The current flowing through a copper voltamet | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The mass deposited $m$ is given by Faraday's law of electrolysis: $m = Z i t$. Also,the mass can be expressed as $m = ho 	imes V = ho 	imes

Vedclass · Accepted Answer

$1.3 	imes 10^{-5}\ m$

Vedclass · Answer

(C) The mass deposited $m$ is given by Faraday's law of electrolysis: $m = Z i t$. Also,the mass can be expressed as $m = ho 	imes V = ho 	imes A 	imes x$,where $ho$ is density,$A$ is area,and $x$ is thickness. Equating the two expressions: $Z i t = ho A x$. Solving for thickness $x$: $x = \frac{Z i t}{A ho}$. Given: $Z = 0.00033\ g/C = 0.00033 	imes 10^{-3}\ kg/C$,$i = 1.5\ A$,$t = 20 	imes 60\ s = 1200\ s$,$A = 50\ cm^2 = 50 	imes 10^{-4}\ m^2$,$ho = 9000\ kg/m^3$. Substituting the values: $x = \frac{0.00033 	imes 10^{-3} 	imes 1.5 	imes 1200}{50 	imes 10^{-4} 	imes 9000}$. $x = \frac{0.00033 	imes 1.5 	imes 1200}{50 	imes 9000} 	imes 10^{-3} = \frac{0.594}{450000} 	imes 10^{-3} = 1.32 	imes 10^{-6} 	imes 10 = 1.3 	imes 10^{-5}\ m$.

The current flowing through a copper voltameter is $1.5\ A$. If the surface area of the electrode plate is $50\ cm^2$,the thickness of the copper layer deposited on this surface in $20\ min$ will be: (Density of copper = $9000\ kg/m^3$ and Electrochemical equivalent of copper = $0.00033\ g/C$)

Similar Questions

$A$ thermocouple uses Bismuth and Tellurium as the dissimilar metals. The sensitivity of Bismuth is $-72\,\mu V/{^oC}$ and that of Tellurium is $500\,\mu V/{^oC}$. If the difference between hot and cold junction is $100\,{^oC}$, then the maximum output will be ............... $mV$.

As the temperature of the hot junction of a thermocouple is increased (while the cold junction is at a constant temperature), the thermo $e.m.f.$

Above neutral temperature,thermo $e.m.f.$ in a thermocouple

The cold junction of a thermocouple is at $0^{\circ} C$. The thermo e.m.f. produced in the thermocouple is given by the equation $E = 16T - 0.04T^2$,where $T$ is the temperature of the hot junction. The temperature of inversion and the neutral temperature of the thermocouple are:

Three identical resistors are connected in series,and the total power consumed is $10 \, W$. If they are connected in parallel,what will be the total power consumed in $W$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module