Find the value of equivalent weight of $$H2C2O4$$ in the following redox reaction.$$K2Cr2O7+H2C2O4$$⟶$$2Cr3++2CO2$$ (at$$.wt., $$K=39$$,$$Cr=52$$,$$O=16$$,$$C=42$$,$$H=1)

Question

Find the value of equivalent weight of  $$H2C2O4$$ in the following redox reaction.$$K2Cr2O7+H2C2O4$$⟶$$2Cr3++2CO2$$ (at$$.wt., $$K=39$$,$$Cr=52$$,$$O=16$$,$$C=42$$,$$H=1)

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Reaction :$$K2Cr2O7+H2C2O4$$⟶$$2Cr3++2CO2(i)$$                    $$Cr2+6+6e$$−⟶$$2Cr3+$$                                      [∵ O.N. of Cr in $$K2Cr2O7=+6$$∴     Eq. wt. of   $$K2Cr2O7=$$  Mol. wt. of $$K2Cr2O7$$ no. of electrons gained by                                                    one molecule of $$K2Cr2O7$$                                                $$=(2p$$×$$39)+(2$$×$$52)+(7$$×$$16)6=2946=49$$ Ans. (ii)                       $$C2+3$$⟶$$2C+4+2e$$−              [∵O.N of C in $$H2C2O4=$$ Mol. wt. of $$H2C2O4$$ no. of electrons lost by                                                           one molecule of $$H2C2O4$$                                                    $$=(2$$×$$1)+(2$$×$$12)+(4$$×$$16)2$$                                                      $$=902=45$$

Find the value of equivalent weight of H2C2O4 in the following redox reaction.K2Cr2O7+H2C2O4⟶2Cr3++2CO2 (at.wt., K=39,Cr=52,O=16,C=42,H=1)

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