\(6 {OH}^{-}+{Cl} ^{-}\rightarrow {ClO}_{3}^{-}+3 {H}_{2} {O}+6 {e}^{-}\)

\(\rightarrow 10\, {~g}\, {KClO}_{3} \Rightarrow \frac{10}{122.6} \,{~mol} \,{KCO} 3\) in obtained

\(\rightarrow\) From the above reaction, it is concluded that by \(6\, {~F}\) charge \(1\, {~mol} \,{KClO}_{3}\) is obtained.

\(\rightarrow\) By the passage of \(6\, {~F}\) charge \(=1\, {~mol} \,{KClO}_{3}\)

\(\therefore\) By the passage of \(\frac{x \times 10 \times 60 \times 60}{96500}\, {~F}\) charge

\(=\frac{1}{6} \times \frac{x \times 10 \times 60 \times 60}{96500}\)

Now \(\frac{x \times 10 \times 60 \times 60}{6 \times 96500}=\frac{10}{122.6}\)

\(\Rightarrow x=\frac{10 \times 965}{60 \times 122.6}=\frac{965}{735.6}=1.311 \simeq 1\)

OR

\(W=\frac{E}{F} \times I \times t\)

\(10=\frac{122.6}{96500 \times 6} \times x \times 10 \times 3600\)

\(X=1.311\)