A `5.0 mL` of solution of `H_(2)O_(2)` liberates `0.508 g` of iodine from acidified `KI` solution. Calculate the strength of `H_(2)O_(2)` solution in terms of volume strength at `STP`.
Text Solution
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The redox reaction involved is : `H_(2)_(2) + 2I^(-) +2H^(+) rarr 2H_(2)O + I_(2)` If M is molarity of `H_(2)O_(2)` solution, then `5M = (0.508 xx 1000)/(254) (because "1 mole" H_(2)O_(2) -= "1 mole" I_(2))` `rArr M=0.4` Also, n-factor of `H_(2)O_(2)` is 2, therefore normality of `H_(2)O_(2)` solution is `0.8 N`. `rArr` Volume strength = Normality `xx 5.6 = 0.8 x 5.6 = 4.48 V`
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