Question

If a current of 4.468 a is passed for 6 hours between nickel electrodes in 0.5 l of 2 m solution of nickel nitrate, what will be the molality of nickel nitrate in solution at the end of electrolysis ? Assuming volume of the solution remains same after electrolysis.

Answer 1

Answer:

The molarity shall remain same as 2M

Explanation:

Let us first consider the chemical reactions at each electrode,

The electrolysis of Ni(NO3)2 in presence of Ni electrode will bring in following changes :

At anode : N i→ Ni2++2e

At cathode : Ni2++2e → Ni

If we compare both the reactions when the electrolysis of Ni(NO3)2 take place, we see that the decomposition of Ni at anode is equally compensated at cathode. Hence, the concentration of Ni(NO3)2 solution remain same during the process

Answer 2

Molality of the given solution at the end of electrolysis is 1.0 m.

Explanation:

The reaction taking place at the electrode is as follows.

$Ni^{2+} + 2e^{-} \rightarrow Ni$

  1 mole      2 × 96500 C

Therefore, the quantity of electric charge passed is calculated as follows.

$4.468 Amp \times 6 \times 60 \times 60$ Coulomb

= 96508.8 C

Hence, moles of $Ni(NO_{3})_{2}$ decomposed or nickel deposited is as follows.

   $\frac{1}{2 \times 96500} \times 96508.8 C$

    = 0.5 C

So, number of moles of $Ni(NO_{3})_{2}] present before hydrolysis are as follows.</p><p>        [tex]0.5 \times 2 = 1.0$

And, moles of $Ni(NO_{3})_{2}] present after the electrolysis are as follows.</p><p>        [tex](1.0 - 0.5) = 0.5$

As 0.5 moles are present in the given solution. Therefore, molality of the solution will be calculated as follows.

         2m  

     = $2 \times 0.5$

     = 1.0 m

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