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

Balance the following redox reaction by ion – electron method :

M n O_{4}^{-} (a q) + I^{-} (a q) \to M n O_{2} (s) + I_{2} (s)

(in basic medium).

Accepted Answer

Step 1: The two half reactions involved in the given reaction are:

Oxidation half reaction: $I^{-} (a q) \to I_{2} (s)$
Reduction half reaction: ${MnO}_{4}^{-} (aq) \to {MnO}_{2} (s)$
Step 2: Balancing $I$ in the oxidation half reaction, we have:
Now, to balance the atom, we multiply $I^{-}$ with $2$ into the left-hand side of the reaction
$2 I^{-} (a q) \to I_{2} (s)$
Now, to balance the charge, we add $2 e^{-}$ to the right-hand side of the reaction.
$2 I^{-} (a q) \to I_{2} (s) + 2 e^{-}$

Step 3: In the reduction half-reaction, Balance the atoms other than O and
H in each half-reaction individually. Here the to balance the atoms add $H_{2} O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms.
Thus, we get : $M n O_{4}^{-} (a q) + 4 H^{+} (aq) \to M n O_{2} (s) + 2 H_{2} O (l)$
Step 4: The reaction occurs in a basic medium, so we add $4 O H^{-}$ ions to the right-hand side and right-hand side.
${MnO}_{4}^{-} (aq) + 4 H^{+} (aq) + 4 {OH}^{-} (aq) \to {MnO}_{2} (s) + 2 H_{2} O (l) + 4 {OH}^{-} (aq)$
= ${MnO}_{4}^{-} (aq) + 2 H_{2} O (l) \to {MnO}_{2} (s)$ + $4 O H^{-} (aq)$
Step 5 : To balance the charge add $3$ electrons are added to the left-hand side of the reaction.
$3 e^{-} + {MnO}_{4}^{-} (aq) + 2 H_{2} O (l) \to {MnO}_{2} (s)$ + $4 O H^{-} (aq)$

Step 6: Equalising the number of electrons by multiplying the oxidation half-reaction by $3$ and the reduction half-reaction by $2$ , we have:
$6 I^{-} (a q) \to {3 I}_{2} (s) + 6 e^{-}$
$2 M n O_{4}^{-} (aq) + 4 H_{2} O (l) + 6 e^{-} \to M n O_{2} (s) + 8 O H^{-} (a q)$
Step 7: Adding the two half-reactions, we have the net balanced redox reaction as:
$6 I^{-} (a q) + 2 M n O_{4}^{-} (a q) + 4 H_{2} O (l) \to 3 I_{2} (s) + 2 M n O_{2} (s) + 8 O H^{-} (a q)$

Balance the following redox reaction by ion – electron method :
$M n O_{4}^{-} (a q) + I^{-} (a q) \to M n O_{2} (s) + I_{2} (s)$ (in basic medium).

Important Questions on Stoichiometry

Learn and Prepare for any exam you want !

Balance the following redox reaction by ion – electron method : MnO4-aq+I-aq→MnO2s+I2s (in basic medium).

Important Questions on Stoichiometry

Learn and Prepare for any exam you want !

Balance the following redox reaction by ion – electron method :
$M n O_{4}^{-} (a q) + I^{-} (a q) \to M n O_{2} (s) + I_{2} (s)$ (in basic medium).