Potassium chlorate is prepared by the electrolusis of `KCl` in basic medium as:
`Cl^(-) + 6OH^(-) rarr ClO_(3)^(-) + 3H_(2)O + 6e`
If only `60%` of current is utilised in the reaction, the time to produce `10 g` of `KClO_(3)` using current of `2` ampere : (`mol. wt.` of `KClO_(3) = 122.5`)

Cr(OH)_(3)+ClO^(-)+3OH^(-)rarr?+Cl^(-)3H_(2)O . The missing ion is

Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation. 6 O H ^(-) + Cl ^(-) to ClO _(3) ^(-) + 3 H _(2) O + 6e^(-) A current of x A has to be passed for 10 h to produce 10.0 g of potassium chlorate. The value of x is ____________ (Nearest integer) (Molar mass of KClO _(3) = 122.6 g mol ^(-1) , F = 96500C)

Complete the following reaction: (i) Cl_(2) + H_(2)O rarr (ii) XeF_(6) + 3H_(2)O rarr

The values of m and n in the following redox reaction, mCl_(2)+6OH^(-) to ClO_(3)^(-)+nCl^(-)+3H_(2)O are

In the reaction Bromine 3Br_2 + 6 OH^(-) rarr 5 Br ^(-) + BrO_3^(-) + 3H_2O