Write the balanced ionic equation for the reaction of zinc with conc. nitric acid to produce zinc nitrate, nitrogen dioxide and water:

Explanation

Step 1: Skeletal equation: `Zn + HNO_3 to Zn^(2+) + NO_2 +H_2O`
Step 2: Assigning oxidation number and identifying atoms that undergo change in O.N.
`overset(0)(Z)n + overset(+1+5-2)(HNO_3) to overset(+2)(Zn)overset(+5 - 2)((NO_3)_2) + overset(+4 - 2)(NO_2) + overset(+1 -2)(H_2O)`
Here, N retains its O.N. in `Zn(NO_3)_2` but changes in `NO_2` from +5 to +4. This means that nitric acid acts both as acid and as oxidising agent. Therefore, the equation is written in the ionic form as:
`overset(0)(Zn) + H^(+) + overset(+5)(NO_3^-) to overset(+2)(Zn) + overset(+4)(NO_2) + H_2O`
Step 3: Equalising O.N. by multiplying Zn by 1 and `NO_3^(-)` and `NO_2` by 2.
`Zn + H^(+) + 2NO_3^(-) to Zn^(2+) + 2NO_2 + H_2O`
Step 4: Step 4: Balancing ionic charges on both sides by adding `3H^(+)` ions to the left side.
`Zn + 4H^+ + 2NO_3^(-) to Zn^(2+) + 2NO_2 + H_2O`
Step 5: Adding one moleculesof H, 0 to the right to obtain the balanced redox equation.
`Zn + 4H^(+) +2NO_3^(-) to Zn^(2+) + 2NO_2 +2H_2O`