# Balance the following redox equation using the half reaction method. Z n + H N O 3 Z n ( N O 3...

## Question:

Balance the following redox equation using the half reaction method.

{eq}Zn+HNO_3 \rightarrow Zn(NO_3)_2+NH_4NO_3+H_2O {/eq}

## Reducing Nature of Alkali Metals:

Alkali metals are strongly reducing in nature due to their lower ionization energies. The reducing power of alkali metals increases from sodium to cesium. Lithium is anomalous in its chemical reactivity and exhibits the strongest reducing power.

The oxidation number of Zn in Zn and {eq}Zn(NO_3)_2 {/eq} are 0 and +2 respectively. Hence, the oxidation number of Zn increases. So, the oxidation half-reaction is given as:

{eq}Zn\rightarrow Zn(NO_3)_2 {/eq}

Balancing {eq}NO_3 {/eq}, we get: {eq}Zn+2NO_3^-\rightarrow Zn(NO_3)_2 {/eq}

Balancing charge, we get: {eq}Zn+2NO_3^-\rightarrow Zn(NO_3)_2+2e^- {/eq}

So, the balanced half-reaction is given as:

{eq}Zn+2NO_3^-\rightarrow Zn(NO_3)_2+2e^- {/eq} ............ (1)

The average oxidation number of N in {eq}HNO_3 {/eq} and {eq}NH_4NO_3 {/eq} are +5 and +1 respectively. Hence, the oxidation number of N decreases. So, the reduction half-reaction is given as:

{eq}HNO_3\rightarrow NH_4NO_3 {/eq}

Balancing N, we get: {eq}2HNO_3\rightarrow NH_4NO_3 {/eq}

Balancing O, we get: {eq}2HNO_3\rightarrow NH_4NO_3+3H_2O {/eq}

Balancing H, we get: {eq}2HNO_3+8H^+\rightarrow NH_4NO_3+3H_2O {/eq}

Balancing charge, we get: {eq}2HNO_3+8H^++8e^-\rightarrow NH_4NO_3+3H_2O {/eq}

So, the balanced reduction half-reaction is given as:

{eq}2HNO_3+8H^++8e^-\rightarrow NH_4NO_3+3H_2O {/eq} .............. (2)

{eq}4\times Equation-(1)+Equation-(2): {/eq}

{eq}4Zn+2HNO_3+8H^++8NO_3^-\rightarrow 4Zn(NO_3)_2+NH_4NO_3+3H_2O {/eq}

Combining {eq}H^+\;and\;NO_3^- {/eq}, we get: {eq}4Zn+10HNO_3\rightarrow 4Zn(NO_3)_2+NH_4NO_3+3H_2O {/eq}

Conclusion:

Balanced redox equation: {eq}4Zn+10HNO_3\rightarrow 4Zn(NO_3)_2+NH_4NO_3+3H_2O {/eq}