To deposit one $gm$ equivalent of an element at an electrode,the quantity of electricity needed is

  • A
    $1 \, \text{ampere}$
  • B
    $96000 \, \text{amperes}$
  • C
    $96500 \, \text{farads}$
  • D
    $96500 \, \text{coulombs}$

Explore More

Similar Questions

How many faraday are needed to reduce a mole of $MnO_4^-$ to $Mn^{2+}$?

Match the column :-
Column $I$ (Reduction process)Column $II$ (Charge required)
$(a)$ $1$ mol of $MnO_4^-$ to $Mn^{2+}$$(p)$ $193000$ $C$
$(b)$ $1$ mol of $Cr_2O_7^{2-}$ to $Cr^{3+}$$(q)$ $289500$ $C$
$(c)$ $1$ mol of $Sn^{4+}$ to $Sn^{2+}$$(r)$ $482500$ $C$
$(d)$ $1$ mol of $Al^{3+}$ to $Al$$(s)$ $579000$ $C$

The atomic weight of $Fe$ is $56$. The weight of $Fe$ deposited from $FeCl_3$ solution by passing $0.6$ Faraday of electricity is $............$ $g$.

The quantity of electricity needed to liberate $0.5 \ g$ equivalent of an element is

$96500 \ C$ of electric charge liberates how many grams of $Cu$ from $CuSO_4$ solution?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo