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Your reaction, $\ce{C6H12O6 -> CO2}$ is a redox half-reaction. The n-factor of a molecule/compound in a redox reaction is defined as the change in the oxidation state per molecule (as defined in Del Pate's answer). The easy to visualize definition is the amount of electrons (in $\pu{mol}$) donate or accepted per $\pu{1 mol}$ of the compound in the given ...

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Instead of asking what n-factor really was, a better question would have been what an equivalent was (But that would have been a duplicate) According to the above question, the formal definition of an equivalent is as follows: An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent ...

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Yes,the definitions are correct. n-factor is required when finding the equivalent weight of the substance. In case of acids equivalent wt.=molecular weight of acid/basicity(no. of replaceable hydrogen atoms) here,the basicity is the n-factor important observation:replaceable atoms of hydrogen are those which are attached to oxygen atoms. And in the same way, ...

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$$\ce{Cu2S + HNO3 -> S + NO2 + Cu(NO3)2 + H2O}$$ Let's balance this reaction using the half-reaction method. The oxidation half reaction (O.H.R) would be: $$\ce{\overset{+1}{Cu}_2\overset{-2}{S} -> 2Cu^2+ + S^0 + 4e-}\tag{O.H.R}\label{ohr}$$ The reduction half reaction (R.H.R) here is: \ce{H\overset{+5}{N}O3 +e- -> \overset{+4}{N}O2} \tag{R.H.R}\...

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