A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode?

Question

A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode? (A) 0.20 (B) 0.05 (C) 0.10 (D) 0.15

Tardigrade Admin · Accepted Answer

0.1 eq. of Ni+2 will be discharged. No. of eq = (No of moles) × (n-factor) 0.1 = (No. of moles) × 2 No. of moles of Ni = (0.1/2) = 0.05

Q. A solution of $N i (N O_{3})_{2}$ is electrolysed between platinum electrodes using $0.1$ Faraday electricity. How many mole of $N i$ will be deposited at the cathode?

0.20

0.05

0.10

0.15

0.1 eq. of $N i^{+ 2}$ will be discharged.
No. of eq = (No of moles) $\times$ (n-factor)
0.1 = (No. of moles) $\times$ 2
No. of moles of Ni = $\frac{0.1}{2} = 0.05$

Q. A solution of Ni(NO3​)2​ is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode?