When a current of $0.1 \, A$ is passed through $200 \, mL$ of a $0.1 \, N$ $AgNO_3$ solution,silver is deposited. How much time in $sec$ is required to remove half of the silver from the solution?

Electricity is passed through an acidic solution of $Cu^{2+}$ until all the $Cu^{2+}$ is exhausted, leading to the deposition of $300 \ mg$ of $Cu$ metal. Subsequently, a current of $600 \ mA$ is passed through the same solution for another $28 \ minutes$ while keeping the total volume of the solution fixed at $200 \ mL$. The total volume of oxygen evolved at $STP$ during the entire process is . . . . . . $mL$. (Nearest integer) [Given: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$, $E_{red}^0 = +0.34 \ V$; $O_{2(g)} + 4H^{+} + 4e^{-} \rightarrow 2H_2O$, $E_{red}^0 = +1.23 \ V$; Molar mass of $Cu = 63.54 \ g \ mol^{-1}$; Faraday Constant $= 96500 \ C \ mol^{-1}$; Molar volume at $STP = 22.4 \ L$]

In two separate experiments,the same quantity of electricity was passed through silver and gold solutions. The amounts of $Ag$ and $Au$ deposited are $2.15 \ g$ and $1.31 \ g$,respectively. The valency of gold is [Atomic mass of $Ag = 107.9$; $Au = 197$].

Two electrolytic cells containing $CuSO_4$ and $AgNO_3$ respectively are connected in series and a current is passed through them until $1 \, mg$ of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately ......... $mg$ [Atomic weights of copper and silver are respectively $63.57$ and $107.88$]

Three faradays of electricity are passed through molten $Al_2O_3$,aqueous solution of $CuSO_4$ and molten $NaCl$ taken in different electrolytic cells. The amount of $Al$,$Cu$ and $Na$ deposited at the cathodes will be in the ratio of

When $9.65 \ C$ of electricity is passed through a solution of silver nitrate (atomic weight of $Ag = 108$),the amount of silver deposited is $............ \ mg$.