Balance the redox reactions by ion – electron method :

$$MnO^{ – }_{ 4 \; (aq) } \; + \; I_{ (aq) }^{ – } \; \rightarrow \; MnO_{ 2 \; (s) } \; + \; I_{ 2 \; (s) }​$$

Asked by Pragya Singh | 1 year ago |  87

##### Solution :-

Step 1

The two half reactions are given below:

Oxidation half reaction: $$I_{ (aq) } \; \rightarrow \; I_{ 2 \; (s) }$$

Reduction half reaction: $$MnO_{ 4 }^{ – } \; \rightarrow \; MnO_{ 2 }$$

Step 2

Balance I in oxidation half reaction:

$$2 \; I^{ – }_{(aq)} \; \rightarrow \; I_{ 2 \; (s) }$$

Add $$2 e^{ – }$$ to the right hand side of the reaction to balance the charge:

$$2I^{ – }_{ (aq) } \; \rightarrow \; I_{ 2 \; (s) } \; + \; 2 \; e^{ – }$$

Step 3

The oxidation no. of Mn has decreased from +7 to +4 in the reduction half reaction. Therefore, 3 electrons are added to the left hand side of the reaction.

$$MnO^{ – }_{ 4 \; (aq) } \; + \; 3 \; e^{ – } \; \rightarrow \; MnO_{ 2 \; (aq)}$$

Add $$4 OH^{ – }$$ ions to right hand side of the reaction to balance the charge.

$$MnO^{ – }_{ 4 \; (aq) } \; + \; 3 \; e^{ – } \; \rightarrow \; MnO_{ 2 \;(aq)} \; + \; 4 \; OH^{ – }$$

Step 4

There are 6 oxygen atoms on the right hand side and 4 oxygen atoms on the left hand side. Hence, 2 water molecules are added to the left hand side.

$$MnO^{ – }_{ 4 \; (aq) } \; + \; 2 \; H_{ 2 }O \; + \; 3\; e^{ – }\; \rightarrow \; MnO_{ 2 \; (aq) } \; + \; 4 OH^{ – }$$

Step 5

Equal the no. of electrons on both the sides by multiplying oxidation half reaction by 3 and reduction half reaction by 2:

$$6 \; I_{ (aq) }^{ – } \; \rightarrow \; 3 \; I_{ 2 \; (s)} \; + \; 6 \; e^{ – }$$

$$2 \; MnO^{ – }_{ 4 \; (aq) } \; + \; 4 \; H_{ 2 }O \; + \; 6\; e^{ – } \;$$

$$\rightarrow 2 \; MnO_{ 2 \; (s)} \; + \; 8\; OH^{ – }_{ (aq) }$$

Step 6

After adding both the half reactions, we get the balanced reaction as given below:

$$6 \; I_{ (aq) }^{ – } \; + \; 2 \; MnO^{ – }_{ 4 \; (aq) } \;+ \; 4 \; H_{ 2 }O_{ (l) } \;$$

$$\rightarrow \; 3 \; I_{ 2 \; (s)} \; + \; 2 \; MnO_{ 2 \; (s)} \; + \; 8\; OH^{ – }_{ (aq) }$$

Answered by Abhisek | 1 year ago

### Related Questions

#### Fluorine reacts with ice and results in the change:

Fluorine reacts with ice and results in the change:

$$H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) }$$

Justify that this reaction is a redox reaction

#### Depict the galvanic cell in which the reaction is

Depict the galvanic cell in which the reaction is:

$$Zn_{\left ( s \right )} \;+ \;2 \;Ag_{\left ( aq \right )}^{+} \;\rightarrow \;Zn^{2+}_{\left ( aq \right )} \;+ \;2 \;Ag_{\left ( s \right )}$$

Further show:

(i) which of the electrode is negatively charged?

(ii) the carriers of the current in the cell.

(iii) individual reaction at each electrode.

#### Given the standard electrode potentials,

Given the standard electrode potentials,

$$\dfrac{ K^{+}}{K}$$= –2.93V

$$\dfrac{ Ag^{+}}{Ag}$$ = 0.80V

$$\dfrac{Hg^{2+}}{Hg}$$ = 0.79V

$$\dfrac{ Mg^{2+}}{Mg}$$ = –2.37V

$$\dfrac{Cr^{3+}}{Cr}$$ = –0.74V

Arrange these metals in their increasing order of reducing power.

#### Arrange the given metals in the order in which they displace each other from the solution of their salts.

Arrange the given metals in the order in which they displace each other from the solution of their salts.

Al, Fe, Cu, Zn, Mg

#### An aqueous solution of AgNO3 with silver electrodes

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of $$AgNO_{3}$$ with silver electrodes

(ii) An aqueous solution $$AgNO_{3}$$​ with platinum electrodes

(iii) A dilute solution of $$H_{2}SO_{4}$$​ with platinum electrodes

(iv) An aqueous solution of $$CuCl_{2}$$with platinum electrodes.