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Standard X

Physics

Resistance

The concentra...

Question

The concentration (in molarity) of Ni(NO $_{3}$ ) $_{2}$ left after passing 965 ampere current for one second through 2M Ni(NO $_{3}$ ) $_{2}$ solution using Ni electrode is:

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Solution

The number of faradays passed $= \frac{965 \times 1}{96500} = 0.01$ .
They will deposit $\frac{0.01}{2} = 0.005$ moles of Ni.
However, the equivalent amount of Ni (from Ni electrode) will dissolve into solution. Hence, the concentration of $N i (N O_{3})_{2}$ will remain unchanged. i.e, it will remain 2 M.

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The concentration (in molarity) of Ni(NO3)2 left after passing 965 ampere current for one second through 2M Ni(NO3)2 solution using Ni electrode is:

The number of faradays passed =965×196500=0.01.They will deposit 0.012=0.005 moles of Ni.However, the equivalent amount of Ni (from Ni electrode) will dissolve into solution. Hence, the concentration of Ni(NO3)2 will remain unchanged. i.e, it will remain 2 M.

The concentration (in molarity) of Ni(NO $_{3}$ ) $_{2}$ left after passing 965 ampere current for one second through 2M Ni(NO $_{3}$ ) $_{2}$ solution using Ni electrode is: