In a reaction, Cr 2 O 72- is reduced to Cr 3+. What will be the concentration of 0.1 M K 2 Cr 2 O 7 in equivalent per litre? Cr 2 O 72-+14 H ++6 e - arrow 2 Cr 3++7 H 2 O

Question

In a reaction, Cr 2 O 72- is reduced to Cr 3+. What will be the concentration of 0.1 M K 2 Cr 2 O 7 in equivalent per litre? Cr 2 O 72-+14 H ++6 e - arrow 2 Cr 3++7 H 2 O (A) 0.9 N (B) 0.6 N (C) 0.3 N (D) 0.2 N

Tardigrade Admin · Accepted Answer

N = M × Valence factor ∵ mathrmN=0.1 × 6(2 mathrmCr6++6 mathrme- arrow 2 mathrmCr3+) K 2 Cr 2 O 7=0.6 N

Q. In a reaction, $C r_{2} O_{7}^{2 -}$ is reduced to $C r^{3 +}$ . What will be the concentration of $0.1 M K_{2} C r_{2} O_{7}$ in equivalent per litre? $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$

$0.9 N$

$0.6 N$

$0.3 N$

$0.2 N$

$N = M \times$ Valence factor
$∵ N = 0.1 \times 6 (2 Cr^{6 +} + 6 e^{-} \to 2 Cr^{3 +})$
$K_{2} C r_{2} O_{7} = 0.6 N$

Q. In a reaction, Cr2​O72−​ is reduced to Cr3+. What will be the concentration of 0.1MK2​Cr2​O7​ in equivalent per litre? Cr2​O72−​+14H++6e−→2Cr3++7H2​O

0.9N

0.6N

0.3N

0.2N