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

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)

The values of X and Y in the following redox reaction are XCl_(2) +6OH rarr ClO_3^(-) +YCl^(-) +3H_(2)O

When electrolysis of KCl is doen in alkaline medium, 10g of KClO_(3) is produced as follows : Cl^(-)+6overset(-)(O)H rarr ClO_(3) +3H_(2)O+6e^(-) A current of 2A is passed for 10.941 hours. Calculate the ((% current efficiency)/(10)) used in the process. (Mw of KClO_(3)=122.5)

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