In hot concentrated alkali iodine disproportionates: 3I2 + 6OH- -> 5I- + IO3- + 3H2O. The n-factor of I2 in this reaction is:

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In hot concentrated alkali iodine disproportionates: $3 I X_{2} + 6 OH X^{-} 5 I X^{-} + IO X_{3} X^{-} + 3 H X_{2} O$ . The n-factor of $I X_{2}$ in this reaction is:

For 3 mol of

I X_{2}

the electrons transferred are 5 (one I goes

0 \to + 5

, losing 5 electrons, and five I go

0 \to - 1

, gaining 5). n-factor per mole of

I X_{2}

= 5/3