For the reaction: `I^(Theta)+ClO_(3)^(Theta)+H_(2)SO_(4)rarrCl^(Theta)+HSO_(4)^(Theta)+I_(2)` The correct statement(s) in the balanced equation is/are
A
stoichiometric coefficient of `HSO_(4)^(-)` is 6
B
iodide is oxidised
C
sulphur is reduced
D
`H_(2)O` is one of the products
Text Solution
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The correct Answer is:
A, B, D
PLAN This problem includes concept of redox reaction.A redox reaction consists of oxidation half-cell reaction and reduction half-cell reaction, Write both half-cell reactions, i.e., oxidation half-cell reaction and reduction half-cell reaction. Then balance both the equations. Now determine the correct value of stoichiometry of `H_(2)SO_(4)`. Oxidation half-reaction, `2I^(-) rarr I_(2) + 2e^(-)` ....(i) Here, `I^(-)` is converted into `I_(2)`. Oxidation number of `I` is increasing from `-1` to 0 hence, this is a type of oxidation reaction. Reduction half-reaction `6H^(+) + ClO_(30^(-) + 6e^(-) rarr Cl^(-) + 3H_(2)O` Here, `H_(2)O` releases as a product. Hence, option (d) is correct. Multiplying equation (i) by 3 and adding in equation (ii) `6I^(-) + ClO_(3)^(-) + 6H^(+) rArr Cl^(-) + 3I_(2)+3H_(2)O` `6I^(-) + ClO_(3)^(-) + 6H_(2)SO_(4) rarr Cl^(-) + 3I_(2)+3H_(2)O+6HSO_(4)^(-)` Stoichiometric coefficient of `HSO_(4)^(-)` is 6. Hence, option (a), (b) and (d) are correct.
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