One mole of acidified K2Cr2O7 on reaction with excess KI will liberate ....... mole(s) of I2 :

Question

One mole of acidified K2Cr2O7 on reaction with excess KI will liberate ....... mole(s) of I2 : (A) 6 (B) 1 (C) 7 (D) 2

Tardigrade Admin · Accepted Answer

Cr2O72-+14H++6e- xrightarrow[]2Cr3++7H2O beginalign underline (2I- xrightarrowI2+2e-)× 3 Cr2O72-+14H++6I- xrightarrow endalign 2Cr3++7H2O+3I2 Hence, number of moles of I2 produced = 3

Q. One mole of acidified $K_{2} C r_{2} O_{7}$ on reaction with excess $K I$ will liberate ....... mole(s) of $I_{2}$ :

6

1

7

2

3

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} 2 C r^{3 +} + 7 H_{2} O$ $\underline{(2 I^{-} I_{2} + 2 e^{-}) \times 3} C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-}$ $2 C r^{3 +} + 7 H_{2} O + 3 I_{2}$ Hence, number of moles of $I_{2}$ produced = 3

Q. One mole of acidified K2​Cr2​O7​ on reaction with excess KI will liberate ....... mole(s) of I2​ :

6

1

7

2

3

Cr2​O72−​+14H++6e−​2Cr3++7H2​O ​(2I−​I2​+2e−)×3​Cr2​O72−​+14H++6I−​​ 2Cr3++7H2​O+3I2​ Hence, number of moles of I2​ produced = 3

Q. One mole of acidified $K_{2} C r_{2} O_{7}$ on reaction with excess $K I$ will liberate ....... mole(s) of $I_{2}$ :

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} 2 C r^{3 +} + 7 H_{2} O$ $\underline{(2 I^{-} I_{2} + 2 e^{-}) \times 3} C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-}$ $2 C r^{3 +} + 7 H_{2} O + 3 I_{2}$ Hence, number of moles of $I_{2}$ produced = 3