Question

What current strength in ampere will be required to liberate 20 g of iodine from potassium
iodide solution in 2 hours?

• A. 4.22
• B. 2.11
• C. 8.25
• D. 10.5

Answer 1

The correct option is B 2.11

Quantity of electricity passed $=Q\left(C\right)=I\left(A\right)\times t\left(s\right)=I\left(A\right)\times 7200s$
The atomic weight of iodine is 254 $\frac{g}{mol}$
Moles of electrons passed $=\frac{Q\left(C\right)}{96500\frac{C}{mol{e}^{-}}}=\frac{I\left(A\right)\times 7200s}{96500\frac{C}{mol{e}^{-}}}$
The weight of iodine deposited $20g=molarmassofiodine\times moleratio\times molesofelectronspassed20g=254\frac{g}{mol}\times \frac{1moliodine}{2mol{e}^{-}}\times \frac{I\left(A\right)\times 7200s}{96500\frac{C}{mol{e}^{-}}}I\left(A\right)=2.11A$