Permanganate $(V I I)$ ion, $M n O_{4}^{-}$ in basic solution oxidises iodide ion, $I^{-}$ to produce molecular iodine $(I_{2})$ and manganese $(I V)$ oxide $(M n O_{2})$ . Write a balanced ionic equation to represent this redox reaction.

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Solution

$Step 1: Skeletal ionic equation$

The skeletal ionic equation is:

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

$Step 2: Two Half Reaction$

$- 1$ $0$
$Oxidation half : I^{-} (a q) \to I_{2} (s)$

$+ 7$ $+ 4$
$Reduction half : M n O_{4}^{-} (a q) \to M n O_{2} (s)$

$Step 3: Balancing First half Reaction$

To balance the $I$ atoms in the oxidation half reaction, we rewrite it as:

$2 I^{-} (a q) \to I_{2} (s)$

$Step 4: Balancing Second half Reaction$

To balance the $O$ atoms in the reduction half reaction, we add two water molecules on the right:

$M n O_{4}^{-} (a q) \to M n O_{2} (s) + 2 H_{2} O (l)$

To balance the $H$ atoms, we add four $H^{+}$ ions on the left:

$M n O_{4}^{-} (a q) + 4 H^{+} (a q) \to M n O_{2} (s) + 2 H_{2} O (l)$

As the reaction takes place in a basic solution, therefore, for four $H^{+}$ ions, we add four $O H^{-}$ ions to both sides of the equation:

$M n O_{4}^{-} (a q) + 4 H^{+} (a q) + 4 O H^{-} (a q) \to M n O_{2} (s) + 2 H_{2} O (l) + 4 O H^{-} (a q)$

Replacing the $H^{+}$ and $O H^{-}$ ions with water, the resultant equation is:

$M n O_{4}^{-} (a q) + 2 H_{2} O (l) \to M n O_{2} (s) + 4 O H^{-} (a q)$

$Step 5: Balancing the charges$

In this step we balance the charges of the two half-reactions in the manner depicted as:

$2 I^{-} (a q) \to I_{2} + 2 e^{-}$

$M n O_{4}^{-} (a q) + 2 H_{2} O (l) + 3 e^{-} \to M n O_{2} (s) + 4 O H^{-} (a q)$

Now to equalise the number of electrons, we multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $2.$

$6 I^{-} (a q) \to 3 I_{2} (s) + 6 e^{-}$

$2 M n O_{4}^{-} (a q) + 4 H_{2} O (l) + 6 e^{-} \to 2 M n O_{2} (s) + 8 O H^{-} (a q)$

$Step 6: Adding two half reactions$

Add two half-reactions to obtain the net reactions after cancelling electrons on both sides.

$6 I^{-} (a q) + 2 M n O_{4}^{-} (a q) + 4 H_{2} O (l) \to 3 I_{2} (s) + 2 M n O_{2} (s) + 8 O H^{-} (a q)$

$Step 7: Final verification of balanced equation$

A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides.

$6 I^{-} (a q) + 2 M n O_{4}^{-} (a q) + 4 H_{2} O (l) \to 3 I_{2} (s) + 2 M n O_{2} (s) + 8 O H^{-} (a q)$

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Q. Permanganate (VII) ion,

M n O_{4}^{-}

oxidies

I^{-}

ion to

I_{2}

and gives manganese (IV) oxide

M n O_{2}

in basic medium. The skeletal ionic equation is given as

p M n {O_{4}}_{(a q)}^{-} + x H_{2} O (l) \to r M n O_{2} (s) + s I_{2} (s) + y O H_{(a q)}^{-}

The values of p,q,r and s are:

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Permanganate(VII) ion, MnO−4 in basic solution oxidises iodide ion, I– to produce molecular iodine (I2) and manganese (IV) oxide (MnO2). Write a balanced ionic equation to represent this redox reaction.

Permanganate $(V I I)$ ion, $M n O_{4}^{-}$ in basic solution oxidises iodide ion, $I^{-}$ to produce molecular iodine $(I_{2})$ and manganese $(I V)$ oxide $(M n O_{2})$ . Write a balanced ionic equation to represent this redox reaction.