Question

Consider the reaction :
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$
What is the quantity of electricity in coulombs needed to reduce $1\,$ mol of $Cr_2O^{2-}_7$ ?

• A. $3 \times 96500 \,C$
• B. $6 \times 96500 \,C$
• C. $2 \times 96500\,C$
• D. $4 \times 96500 \,C$

Answer 1

Answer: (B)

1 mole of $Cr_2O^{2-}_7$ requires 6 moles of electrons for reduction.
$\therefore $ Required charge $= 6 \times$ Faraday $= 6 \times 96500 \,C$