Name: Redox Reactions as a Basis for Titration
Uploaded: 2019-07-19
Duration: 7 min 17 s
Description: This video will help us learn about how redox indicators like potassium permanganate or diphenylamine are used to indicate the completion of a redox reaction...

Question

Write the balanced ionic equation which represents the oxidation of iodide (

I^{-}

) ion by permanganate ion in basic medium to give iodine (

I

) and manganese dioxide (

{MnO}_{2}

).

Accepted Answer

Step $1$ : Write the unbalanced (skeletal) ionic equation :
${MnO}_{4}^{-} (aq) + I^{-} (aq) \to {MnO}_{2} (s) + I_{2} (s)$
Step $2$ : Assign the oxidation numbers :
$\overset{+ 7 - 2}{{MnO}_{4}^{-}} (aq) + \overset{- 1}{I^{-}} (aq) \to \overset{+ 4 - 2}{{MnO}_{2}} (s) + \overset{0}{I_{2}} (s)$
Step $3$ : Identify the oxidising and reducing agents to construct the unbalanced half-reaction :
Oxidation half-reaction :
$\overset{- 1}{I^{-}} (aq) \to \overset{0}{I_{2}} (s)$
Reduction half-reaction:
$\overset{+ 7 - 2}{{MnO}_{4}^{-}} (aq) \to \overset{+ 4 - 2}{{MnO}_{2}} (s)$

Step $4$ : (a) To balance the $I$ atoms in the oxidation half-reaction, we write
$2 I^{-} (aq) \to I_{2} (s)$
(b) To account for the change in oxidation numbers, we add two electrons to the right-hand side fo oxidation half-reaction and three electrons to the left-hand side of reduction half-reaction:
$2 I^{-} (aq) \to I_{2} (s) + 2 e^{-}$
${MnO}_{4}^{-} (aq) + 3 e^{-} \to {MnO}_{2} (s)$
( 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 :
${MnO}_{4}^{-} (aq) + 3 e^{-} \to {MnO}_{2} (s) + 4 {OH}^{-} (aq)$
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.
${MnO}_{4}^{-} (aq) + 2 H_{2} O + 3 e^{-} \to {MnO}_{2} (s) + 4 {OH}^{-} (aq)$

Step $5$ : Balance the electron transfer. To equalise the number of electrons, we multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $2$ :
$6 I^{-} (aq) \to 3 I_{2} (s) + 6 e^{-} 2 {MnO}_{4}^{-} (aq) + 4 H_{2} O + 6 e^{-} \to 2 {MnO}_{2} (s) + 8 {OH}^{-} (aq)$
Step $6$ : Add the two half-reactions to give, after cancelling the electrons on both sides, we get,
$2 {MnO}_{4}^{-} (aq) + 4 H_{2} O + 6 I^{-} (aq) \to 2 {MnO}_{2} (s) + 3 I_{2} (s) + 8 {OH}^{-} (aq)$
Step $7$ : A final check shows that the equation is balanced with respect to both atoms and charges.

Write the balanced ionic equation which represents the oxidation of iodide ( $I^{-}$ ) ion by permanganate ion in basic medium to give iodine ( $I$ ) and manganese dioxide ( ${MnO}_{2}$ ).

Important Questions on Stoichiometry

Learn and Prepare for any exam you want !

Write the balanced ionic equation which represents the oxidation of iodide (I-) ion by permanganate ion in basic medium to give iodine (I) and manganese dioxide (MnO2).

Important Questions on Stoichiometry

Learn and Prepare for any exam you want !

Write the balanced ionic equation which represents the oxidation of iodide ( $I^{-}$ ) ion by permanganate ion in basic medium to give iodine ( $I$ ) and manganese dioxide ( ${MnO}_{2}$ ).