Passage of one ampere current through 0.1 M Ni(NO₃)₂ solution using Ni electrodes bring in the concentration of solution to _ in 60 seconds.(round off the answer to single digit after decimal)

Solution:

The cell reactions taking place at the electrodes are :

At cathode : Ni²⁺ + 2e $\to$Ni

At anode    : Ni $\to$ Ni²⁺ + 2e 

Charge required to deposit 1 mole of Ni is $2\times 96500 \mathrm{C}$, but the charge passed is:

$\mathrm{Q}=\mathrm{I}\times \mathrm{t}=1\times 60=60 \mathrm{C}$

Mass of Nickel deposited is calculated as:

$\mathrm{W}=\mathrm{ZIt}=\frac{\mathrm{E}}{96500}\times \mathrm{I}\times \mathrm{t}$

Substitute the values:

$\mathrm{W}=\frac{{\displaystyle \frac{58.69}{2}}}{96500}\times 1\times 60=0.0182 \mathrm{g}$

The number of moles deposited:

 $\mathrm{Moles}=\frac{0.0182 \mathrm{g}}{58.6 \mathrm{g}/\mathrm{mol}}=3.1\times {10}^{-4} \mathrm{mol}$

0.1 M Ni(NO₃)₂ indicates that there is 0.1 Ni²⁺ moles and the number of moles of Ni²⁺ remaining in 1 L solution:

$\mathrm{Moles} \mathrm{remaining}=\left(0.1-3.1\times {10}^{-4}\right) \mathrm{mol}=0.09969 \mathrm{mol}$

$\mathrm{Molarity}=\frac{0.09960 \mathrm{mol}}{1 \mathrm{L}}=0.09960 \mathrm{M}\cong 0.1 \mathrm{M}$