JEE Advance - Chemistry (2020 - Paper 2 Offline - No. 3)
In the chemical reaction between stoichiometric quantities of KMnO4 and KI in weakly basic solution, what is the number of moles of I2 released for 4 moles of KMnO4 consumed?
Answer
6
Explanation
In alkaline medium : Iodide is oxidised to iodate
$$2MnO_4^ - + {H_2}O + {I^ - }\buildrel {} \over \longrightarrow 2Mn{O_2} + 2O{H^ - } + IO_3^ - $$
But in weakly basic solution :
$$\mathop {KMn{O_4}}\limits^{ + 7} + \mathop {KI}\limits^{ - 1} \buildrel {} \over \longrightarrow \mathop {Mn{O_2}}\limits^{ + 4} + \mathop {{I_2}}\limits^0 $$
Eq. of KMnO4 = Eq. of I2
$$4 \times 3 = n \times 2 \Rightarrow n = 6$$
$$2MnO_4^ - + {H_2}O + {I^ - }\buildrel {} \over \longrightarrow 2Mn{O_2} + 2O{H^ - } + IO_3^ - $$
But in weakly basic solution :
$$\mathop {KMn{O_4}}\limits^{ + 7} + \mathop {KI}\limits^{ - 1} \buildrel {} \over \longrightarrow \mathop {Mn{O_2}}\limits^{ + 4} + \mathop {{I_2}}\limits^0 $$
Eq. of KMnO4 = Eq. of I2
$$4 \times 3 = n \times 2 \Rightarrow n = 6$$
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