Question

c. Calculate the mass of lead that will be deposited from an aqueous solution of lead (II) salt by the same quantity of electricity that deposits 2.70g of copper from an aqueous solution of copper (II) salt. (Cu= 63.5g mol-1; Pb = 207g mol-1).​

Answer 1

Answer:

First, we'll look at formula for Molarity.

Molarity = No. of moles of solute / Vol. of solution (in L)

Before putting values, it helps to know that 50mL=0.05L

Now, 0.1M=n0.05L where n are the no. of moles of Pb(NO3)2

n=0.005 moles

Now we also know that n= given mass / molar mass

Molar mass of Pb(NO3)2=207+2(14+3(16))=331gmmol

Given mass =331gmmol×0.005mol

So mass of Pb(NO3)2=1.655gm

Explanation:

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Answer 2

Given:  2.70 g of Copper is deposited by a certain amount of current when it is passed through the Cu(II) salt.

           Molar mass of Cu = 63.5 g

           Molar Mass of Pb = 207 g

To Find : Mass of Lead deposited when the same amount of current is passed through a Pb(II) salt solution.

           

Solution:

• According to Faraday's Law of Electrolysis,

The masses of different ions liberated at the electrodes, when the same amount of electricity is passed through different electrolytes are directly proportional to their chemical equivalents.

Mathematically,

Mass of element deposited on electrode ∝ equivalent mass of element

                                  W ∝ $\frac{M}{n}$

where, W = weight of element deposited

            M = molecular mass of element

            n = n-factor

• Using Faraday's Law, we will calculate mass of Lead deposited.

1. Mass of Cu deposited = 2.70 g

    2.Equivalent weight of Cu(II) = $\frac{Molecular Mass}{n-factor}$ = $\frac{63.5}{2}$

    3.Let mass of Pb deposited be x g

    4. Equivalent weight of Pb(II) = $\frac{207}{2}$

Now, substituting these values in (1)

                         ⇒ $\frac{x}{2.7}$ = $\frac{\frac{207}{2} }{\frac{63.5}{2} }$

                         ⇒ $\frac{x}{2.7}$ = $\frac{207}{63.5}$                              ( 2 gets cancelled )

                        ⇒ x = $\frac{207}{63.5}$ × 2.7 = 8.8 g

Thus, 8.8 g of Lead is deposited on the electrode from its Pb(II) solution.