C2O−27+I−+H+→Cr+3+l2+H2O The equivalent weight of the reductant in the above equation is :- (At.wt. of Cr=52,I=127)
A
26
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B
127
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C
63.5
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D
36
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Solution
The correct option is D36 +6Cr2O2−7⟶2Cr3+⇒Reduction Cr2O2−7⟶2Cr3++7H2O Cr2O2−7+14H+⟶2Cr3++7H2O ⟹Cr2O2−7+14H++6e−⟶2Cr3++7H2O ⟹(2I−⟶I2+2e−)3 ⟹6I−⟶3I2+6e− So, equivalent weight of Cr2O2−7=2166=36