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_2 O`
Step 2: Assigning oxidation number and identifying atoms that undergo change in O.N.
`overset(0)(Zn) + overset(+1+5-2) (HNO_(3)) to overset(+2+5 -2)(Zn(NO_3)_2) + overset(+4-2)(NO_2) + overset(+1-2)(H_2 O)`
Here, N retains its O.N. in Zn(NO), but changes in NO, 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^(+)) + H^(+) + overset(+5)(NO_(3)^(-)) to overset(+2)(Zn )+ overet(+4)(NO_2) +H_2 O`
Step 3: Equalising O.N. by multiplying Zn by 1 and `NO_(3)^(-)` and `NO_2` by 2.
` Zn + 4H^(+) + 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_2 O`
Step 5: Adding one molecules of `H_2 O` to the right to obtain the balanced redox equation.
` Zn + 4H^(+) + 2NO_(3)^(-) to Zn^(2+) + 2NO_(2) + 2H_2 O`