Question

A vanadium electrode is oxidized electrically, if the magnetic moment (spin only) of the vanadium product is $1.73BM$ and the mass of electrode decreases by $102mg$ 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 $=1.73BM=\sqrt{3}$

Number of unpaired electron in vanadium will be 1 ,

so it confirms that $V$ will exist as

$V^{4+}$ in its compound

$102mg=\frac{102}{51}\times 4m_{eq}$

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

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

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