Question

A vanadium electrode is oxidized electrically, if the magnetic moment (spin only) of the vanadium product is $1.73 \,BM$ and the mass of electrode decreases by $102 \,mg$ during the process, then what will be the amount of current in ampere if it flowed for $8$ seconds? (Assume current efficiency $100\%, V = 51)$

• A. 96.5
• B. 500
• C. 38.6
• D. None

Answer 1

Answer: (A)

Since magnetic moment $=1.73 BM =\sqrt{3}$

Number of unpaired electron in vanadium will be 1 ,

so it confirms that $V$ will exist as

$V ^{4+}$ in its compound

$102 \,mg =\frac{102}{51} \times 4 m _{ eq }$

$\frac{102}{51} \times 4 \times 96,500 \times 10^{-3} C =2 \times 4 \times 96.5 C$

$i \times t=Q,$ where $Q$ is the amount of charge.

$i=\frac{2 \times 4 \times 96.5}{8}=96.5$