For the balanced redox reaction aZn + b , HNO | vedclass

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Question

(C) The oxidation half-reaction is: $Zn 	o Zn^{2+} + 2e^-$. The reduction half-reaction is: $NO_3^- + 10H^+ + 8e^- 	o NH_4^+ + 3H_2O$. To balance el

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$10$

Vedclass · Answer

(C) The oxidation half-reaction is: $Zn 	o Zn^{2+} + 2e^-$. The reduction half-reaction is: $NO_3^- + 10H^+ + 8e^- 	o NH_4^+ + 3H_2O$. To balance electrons,multiply the oxidation half-reaction by $4$: $4Zn 	o 4Zn^{2+} + 8e^-$. Adding the two half-reactions: $4Zn + NO_3^- + 10H^+ 	o 4Zn^{2+} + NH_4^+ + 3H_2O$. Adding nitrate ions to both sides to balance the equation: $4Zn + 10HNO_3 	o 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O$. Comparing this with the given equation $aZn + b \, HNO_3 	o c \, Zn(NO_3)_2 + d \, NH_4NO_3 + 3H_2O$,we get $b = 10$.

For the balanced redox reaction $aZn + b \, HNO_3 \to c \, Zn(NO_3)_2 + d \, NH_4NO_3 + 3H_2O$,the value of $b$ is:

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