CuSO 4 (aq) + Zn (s) → Cu(s) + ZnSO 4 (aq) · Part 12

to represent this redox reaction. Step : First we write the skeletal ionic equation, which is – (aq) + I – (aq) → MnO (s) + I (s) Step : The two half-reactions are: – Oxidation half : I – (aq) → I (s) + + Reduction half: MnO – (aq) → MnO (s) Step : To balance the I atoms in the oxidation half reaction, we rewrite it as: 2I – (aq) → I (s) Step : To balance the O atoms in the reduction half reaction, we add two water molecules on the right: – (aq) → MnO (s) + H O (l) To balance the H atoms, we add four H + ions on the left: – (aq) + H + (aq) → MnO (s) + 2H O (l) As the reaction takes place in a basic solution, therefore, for four H + ions, we add four OH – ions to both sides of the equation: – (aq) + 4H + (aq) + 4OH – (aq) → MnO (s) + H O(l) + 4OH – (aq) Replacing the H + and OH – ions with water, the resultant equation is: – (aq) + 2H O (l) → MnO (s) + OH – (aq) Step : In this step we balance the charges of the two half-reactions in the manner depicted as: 2I – (aq) → I (s) + 2e – – (aq) + 2H O(l) + 3e – → MnO (s) + 4OH – (aq) Now to equalise the number of electrons, we multiply the oxidation half-reaction by and the reduction half-reaction by . 6I – (aq) → 3I (s) + 6e – MnO – (aq) + 4H O (l) +6e – → 2MnO (s) + 8OH – (aq) Step : Add two half-reactions to obtain the net reactions after cancelling electrons on both sides. 6I – (aq) + 2MnO – (aq) + 4H O(l)