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**Hint:**We need to find the equivalent weight of $\text{C}{{\text{l}}_{2}}$ and ${{\text{H}}_{2}}$ to apply second law of Faraday. This electrolysis’ law states that when electricity is passed through electrolyte then, the mass of the substance deposited is proportional to its respective equivalent weight. Mathematically, it is represented as: $\text{mass deposited= Zit or Zq}$

**Complete answer:**Let us solve this question using second law equation and the reaction given:

Step(1)- In order to find Z, we need molecular mass and n-factor of both the gases.

As,$\text{Z = }\dfrac{\text{molecular mass}}{\text{n-factor}\times 96500}$. N-factor of $\text{C}{{\text{l}}_{2}}$will be 2 because the change in oxidation state is 2.

$\text{C}{{\text{l}}^{^{-}}}$ion from$\text{KCl}$ has oxidation state as -1 and $\text{C}{{\text{l}}_{2}}$ has 0 oxidation state.

The change will be 1 but due to presence of 2 mole ions, the change in oxidation state will be 2; $\left( \begin{align}

& \text{2C}{{\text{l}}^{-}}\to \text{C}{{\text{l}}_{2}} \\

& \text{-1}\times \text{2 0} \\

& \text{ -2 0} \\

\end{align} \right)$.

Similarly with${{\text{H}}^{+}}$ exist in water in +1 state and in hydrogen gas as 0. The change will in oxidation state will be 2,

$\left( \begin{align}

& \text{2}{{\text{H}}^{+}}\to {{\text{H}}_{2}} \\

& \text{+1}\times \text{2 0} \\

& \text{ +2 0} \\

\end{align} \right)$.

The molecular mass of $\text{C}{{\text{l}}_{2}}$ and ${{\text{H}}_{2}}$is 71 grams and 2 grams.

Step (2)- Apply the formula to find the weight of ${{\text{H}}_{2}}$ and $\text{C}{{\text{l}}_{2}}$:

q is given as 10000 C,

${{\text{(mass deposited)}}_{{{\text{H}}_{2}}}}\text{= }\dfrac{1\text{ gram }\times \text{10000}}{96500}=\frac{10000}{96500}=0.103\text{ grams}$ and${{\text{(mass deposited)}}_{\text{C}{{\text{l}}_{2}}}}\text{= }\dfrac{71\times 10000}{2\times 96500}=\dfrac{35.5\times 10000}{96500}=0.367\text{ grams}$.

Thus the mass deposited of ${{\text{H}}_{2}}$ and $\text{C}{{\text{l}}_{2}}$ are 0.103 and 0.367 grams.

Step (3)- Find the volume of the gases at NTP; to find that we have to use ideal gas equation, $\text{PV=nRT}$; at NTP, temperature is 293 K, pressure is 1 atm, R is 0.0821 and moles are $\dfrac{\text{weight of gas}}{\text{molar mass}}$.

Moles of $\text{C}{{\text{l}}_{2}}$ are $\dfrac{\text{weight of C}{{\text{l}}_{2}}}{\text{molar mass}}=\dfrac{0.367}{71}=0.0051$

and of ${{\text{H}}_{2}}$are$\dfrac{\text{weight of }{{\text{H}}_{2}}}{\text{molar mass}}=\dfrac{0.103}{2}=0.0515$ .

The volume will be $1\times {{\text{V}}_{\text{C}{{\text{l}}_{2}}}}=0.0051\times 293\times 0.0821$ and $1\times {{\text{V}}_{{{\text{H}}_{2}}}}=0.0515\times 293\times 0.0821$.

The volume of $\text{C}{{\text{l}}_{2}}$and ${{\text{H}}_{2}}$are 0.122 L and 1.23 L.

- The answer to the question is the volume of $\text{C}{{\text{l}}_{2}}$ and ${{\text{H}}_{2}}$ are 0.122 L and 1.23 L and the mass deposited of ${{\text{H}}_{2}}$ and $\text{C}{{\text{l}}_{2}}$ are 0.103 and 0.367 grams.

**Note:**Do not consider Temperature as 273K, while solving the ideal gas equation. As, the conditions of NTP and conditions of STP are different. Temperature and pressure of STP are 273K and 1 atm. The temperature and pressure of NTP are 293K and 1 atm.

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