Question

n-factor of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is

Answer 1

Step 1: Reaction involved :

${\mathrm{K}}_2{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{I}}^{-}\to 2{\mathrm{K}}^{+}+2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}+3{\mathrm{I}}_2$

Here, the n-factor is defined according to the number of moles that would be exchanged for 1 mole of salt.

Step 2: Finding the n-factor:

As, the molecular weight of Potassium dichromate = $294\mathrm{g}/\mathrm{mol}$

There would be an overall gain of 6 electrons

Therefore, the n-factor of the reaction ${\mathrm{K}}_2{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{I}}^{-}\to 2{\mathrm{K}}^{+}+2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}+3{\mathrm{I}}_2$ is 6