step- we can notice that there is change in oxidation state of chromium and sulphur. Oxidation state of chromium changes form + to + . There is decrease of + in oxidation state of chromium on right hand side of the equation. Oxidation state of sulphur changes from + to + .
There is an increase of + in the oxidation state of sulphur on right hand side. To make the increase and decrease of oxidation state equal, place numeral before cromium ion on right hand side and numeral before sulphate ion on right hand side and balance the chromium and sulphur atoms on both the sides of the equation. Thus we get + – + – + – (aq) + 3SO – (aq) → 2Cr + (aq) + + – 3SO – (aq) Step : As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add 8H + on the left to make ionic charges equal – (aq) + 3SO – (aq)+ 8H + → 2Cr + (aq) + 3SO – (aq) Step : Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., 4H O) on the right to achieve balanced redox change. – (aq) + 3SO – (aq)+ 8H + (aq) → 2Cr + (aq) + 3SO – (aq) +4H O (l) Problem .
Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction. Step : The skeletal ionic equation is : – (aq) + Br – (aq) → MnO (s) + BrO – (aq) Step : Assign oxidation numbers for Mn and Br + – + + – (aq) + Br – (aq) → MnO (s) + BrO – (aq) this indicates that permanganate ion is the oxidant and bromide ion is the reductant. Step : Calculate the increase and decrease of oxidation number, and make the increase equal to the decrease.