Consider the reaction: Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O2-7 ?

Question

Consider the reaction: Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O2-7 ? (A) 3 × 96500 C (B) 6 × 96500 C (C) 2 × 96500 C (D) 4 × 96500 C

Tardigrade Admin · Accepted Answer

1 mole of Cr2O2-7 requires 6 moles of electrons for reduction. ∴ Required charge = 6 × Faraday = 6 × 96500 C

Q. 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 $C r_{2} O_{7}^{2 -}$ ?

$3 \times 96500 C$

$6 \times 96500 C$

$2 \times 96500 C$

$4 \times 96500 C$

1 mole of $C r_{2} O_{7}^{2 -}$ requires 6 moles of electrons for reduction.
$∴$ Required charge $= 6 \times$ Faraday $= 6 \times 96500 C$

Q. Consider the reaction : Cr2​O72−​+14H++6e−→2Cr3++7H2​O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2​O72−​ ?

3×96500C

6×96500C

2×96500C

4×96500C