Consider the reaction, textCr2 textO72 - + 1 4 textH+ + 6 texte- arrow 2 textCr3 + + 7 textH2 textO What is the quantity of electricity in coulombs needed to reduce 1 mole of textCr2 textO72 - ? (Given: 1F ≈ 96500 C)

Question

Consider the reaction, textCr2 textO72 - + 1 4 textH+ + 6 texte- arrow 2 textCr3 + + 7 textH2 textO What is the quantity of electricity in coulombs needed to reduce 1 mole of textCr2 textO72 - ? (Given: 1F &asymp; 96500 C) (A) 5.79 x 105 (B) 5.69 x 105 (C) 5.59 x 105 (D) 5.49 x 105

Tardigrade Admin · Accepted Answer

1 mole textCr2 textO72 - require 6 moles of electrons to completely reduce into Cr3 + . That means 6F electricity is required. (One Faraday = 96500C) Hence, the charge in coulombs =6x96500C=5.79x105C

Q. Consider the reaction,
$Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O$
What is the quantity of electricity in coulombs needed to reduce 1 mole of $Cr_{2} O_{7}^{2 -}$ ?
(Given: 1F ≈ 96500 C)

5.79 x 10⁵

5.69 x 10⁵

5.59 x 10⁵

5.49 x 10⁵

1 mole $Cr_{2} O_{7}^{2 -}$ require 6 moles of electrons to completely reduce into $C r^{3 +}$ . That means 6F electricity is required. (One Faraday = 96500C)
Hence, the charge in coulombs $= 6 x 96500 C = 5.79 x 1 0^{5} C$

Q. Consider the reaction, Cr2​O72−​+14H++6e−→2Cr3++7H2​O What is the quantity of electricity in coulombs needed to reduce 1 mole of Cr2​O72−​ ? (Given: 1F ≈ 96500 C)

5.79 x 105

5.69 x 105

5.59 x 105

5.49 x 105