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

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}