Question

Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.

6OH^– + Cl^– →  ClO₃ ^– + 3H₂O + 6e^-

A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. the value of x is ..... (Nearest integer)

(Molar mass of KClO₃ = 122.6 g mol^-1 F = 96500 C)

Answer 1

Answer is : 1

Given balanced equation is

→ 10g KClO₃ ⇒ \(\frac{10}{122.6}\) mol KCO₃ in obtained

→ from the above reaction, it is concluded that by 6F charge 1 mol KCIO₃ is obtained.

→ By the passage of 6 F charge = 1 mol KCIO₃

∴ By the Passage of  \(\frac{x\times10\times60 \times 60}{96500}\) F charge

= \(\frac{1}{6}\times\frac{x\times10\times60 \times 60}{96500}\)  

Now \(\frac{x\times10\times60 \times 60}{6\times96500}\) = \(\frac{10}{12.6}\) 

x = \(\frac{10\times965}{6\times122.6}\) = \(\frac{965}{735.6}\) = 1.311 = 1

OR 

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