If $63.5 \ g$ of $Cu$ is deposited on the electrode from a $CuSO_4$ solution,what is the number of electrons involved?

  • A
    $6.022 \times 10^{23}$
  • B
    $3.011 \times 10^{23}$
  • C
    $12.044 \times 10^{23}$
  • D
    $6.022 \times 10^{22}$

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Similar Questions

During electrolysis of aqueous $NaOH$,$4 \ g$ of $O_2$ gas is liberated at $NTP$ at the anode. The volume of $H_2$ gas liberated at the cathode in $litres$ is $..............$

$A$ and $B$ are two metals. The standard reduction potentials of $A^{+}_{(aq)} / A_{(s)}$ and $B^{+}_{(aq)} / B_{(s)}$ are $-0.5 \ V$ and $+0.5 \ V$ respectively. What is the $\log K_C$ value for the following reaction at $298 \ K$?
$A_{(s)} + B^{+}_{(aq)} \rightleftharpoons A^{+}_{(aq)} + B_{(s)}$
(Given: $\frac{2.303 RT}{F} = 0.06 \ V$)

An aqueous solution containing $6.5 \ g$ of $NaCl$ of $90 \%$ purity was subjected to electrolysis. After the complete electrolysis,the solution was evaporated to get solid $NaOH$. The volume of $1 \ M$ acetic acid required to neutralize $NaOH$ obtained above is (in $cm^{3}$)

Consider the following statements pertaining to fuel cells :-
$(a)$ Hydrogen-oxygen fuel cells make use of concentrated $KOH$ solution as an electrolyte and porous graphite impregnated with platinum as electrodes.
$(b)$ The efficiency of a fuel cell is less than unity due to polarization at electrodes and the resistance offered by the electrode and the electrolyte.
$(c)$ The electrical work,assuming the cell to be working reversibly,may be represented as $-\Delta G = W_{\text{electrical}} = -\Delta H + T\Delta S$.
Which of the above statements are correct?

Identify the correct statements from the following:
$(A)$ At $298 \ K$,the potential of a hydrogen electrode placed in a solution of $pH = 10$ is $-0.59 \ V$.
$(B)$ The limiting molar conductivity of $Ca^{2+}$ and $Cl^{-}$ are $119$ and $76 \ S \ cm^2 \ mol^{-1}$ respectively. The limiting molar conductivity of $CaCl_2$ is $195 \ S \ cm^2 \ mol^{-1}$.
$(C)$ The correct relationship between $K_{c}$ and $E_{cell}^{0}$ is $E_{cell}^{0} = \frac{2.303 RT}{nF} \log K_{c}$.

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