How many Faradays are required to reduce 1 mol of Cr 2 O 72- to Cr 3+ in acidic medium?

Question

How many Faradays are required to reduce 1 mol of Cr 2 O 72- to Cr 3+ in acidic medium? (A) 2 (B) 3 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

In acidic medium, Cr 2 O 72-+14 H ++6 e- arrow 2 Cr 3++7 H 2 O Thus, n=6 Hence, 6 F charge is required to reduce 1 mol of Cr 2 O 72- to Cr 3+ in acidic medium.

Q. How many Faradays are required to reduce $1$ mol of $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ in acidic medium?

2

3

5

6

In acidic medium,

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

Thus, $n = 6$

Hence, $6 F$ charge is required to reduce $1 m o l$ of $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ in acidic medium.

Q. How many Faradays are required to reduce 1 mol of Cr2​O72−​ to Cr3+ in acidic medium?

2

3

5

6

In acidic medium, Cr2​O72−​+14H++6e−→2Cr3++7H2​O Thus, n=6 Hence, 6F charge is required to reduce 1mol of Cr2​O72−​ to Cr3+ in acidic medium.

Q. How many Faradays are required to reduce $1$ mol of $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ in acidic medium?

In acidic medium,

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

Thus, $n = 6$

Hence, $6 F$ charge is required to reduce $1 m o l$ of $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ in acidic medium.