Question

Write a balanced ionic equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO-4) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2).

Use the 7 step

Answer 1

step 1: write the unbalanced ionic equation

$MnO_4^-+I^- \to MnO_2+I_2$

step 2: assign the oxidation numbers

$MnO_2^-+I^- \to MnO_2+I_2$

on the left hand side Mn has (-7) , O has (-2), and I$^-$ has (-1).

on the right hand side, Mn has (+4) , O has (-2), and I$_2$ has (0).

step 3: Identify the oxidizing and reducing agents to construct the unbalanced half-reaction :

Oxidation half-reaction

$I^- \to I_2$

Reduction half-reaction:

$MnO_4^- \to MnO_2$

Step 4: (a) To balance the I atoms in the oxidation half-reaction, we write

$2I^- \to I_2$

(b) To account for the change in oxidation numbers, we add two electrons to the right-hand side for oxidation half-reaction and three electrons to the left-hand side of reduction half-reaction:

$2I^-\to I_2+2e^-$

$MnO_2+3e^-\to MnO_2$

( c ) Balance the ionic charges in each half-reaction by adding OH$^-$ ions as the reaction carried out in basic medium. There is no need of this in oxidation half-reaction. In reduction half-reaction we add four OH$^-$ ions to the right -hand side :

$MnO_4^-+3e^-\to MnO_2+4OH^-$

Add H$_2$ O molecules to balance the H and O atoms . There is no such requirement in oxidation half-reaction. In reduction half-reaction, we add two H$_2$ O molecules on the left.

$MnO_4^-+2H_2O+3e^-\to MnO_2+4OH^-$

Step 5: Balance the electron transfer. To equalize the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:

$6I^-\to 3I_2+6e^-$

$2MnO_4^-+4H_2O+6e^-\to 2MnO_2+8OH^-$

Step 6: Add the two half-reactions to give

$MnO_4+6I^-+4H_2O+6e^-\to 2MnO_2+3I_2+8OH^-+6e^-$

After cancelling the electrons on both sides, we obtain

$MnO_4+6I^-+4H_2O+\to 2MnO_2+3I_2+8OH^-$

Step 7: A final check shows that the equation is balanced with respect to both atoms and charges.