Consider the reaction : $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-}$ $\to$ $2 C r^{3 +} + 7 H_{2} O$
What is the quantity of electricity in coulombs needed to reduce 1 mol of $C r_{2} O_{7}^{2 -}$ ?

Open in App

Solution

According to equation for the reduction of one mole of $C r_{2} O_{7}^{2 -}$ , 6 mole of electrons are needed.

Hence the amount of electricity require = 6F

= 6 x 96500 C = 579000 C

Suggest Corrections

Similar questions

What is the quantity of electricity in coulombs needed to reduce 1 mol of

? Consider the reaction:

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O

C r_{2} O_{7}^{2 -}

{C r_{2} O_{7}}^{2 -}

M

{C r_{2} O_{7}}^{2 -} + 6 I^{-} + 14 H^{+} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O

C r_{2} O_{7}^{2 -} (a q) + 14 H^{\oplus} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O

C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e \to 2 C r^{3 +} (a q) + 7 H_{2} O

Join BYJU'S Learning Program