Question

Cu + HNO3 ---- Cu(NO3)2 + NO + H2O balance by ion - electron method

Answer 1

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Answer 2

Ion-electron method is one of the ways of balancing equations. In this method, the equation is divided into two parts, one for oxidation and one for reduction. Each of these parts is balanced and then is combined to give a balanced chemical equation. The ion-electron method is also known as half-reaction method.

$\mathrm{Cu}+\mathrm{HNO}_{3} \rightarrow\left(\mathrm{CuNO}_{3}\right)_{2}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}$

First, assign oxidation numbers to each element of the reactants and the products.

$\begin{matrix} 0  \\ Cu \end{matrix} \ \ + \ \ \begin{matrix} +1+5-2 \\ HN{ O }_{ 3 } \end{matrix} \ \ \rightarrow \ \ \begin{matrix} +2+5-2 \\ { \left( Cu{ NO }_{ 3 } \right)  }_{ 2 } \end{matrix} \ \ + \ \ \begin{matrix} +2-2 \\ NO \end{matrix} \ \ + \ \ \begin{matrix} +1-2 \\ { H }_{ 2 }O \end{matrix}$

Now, separate the equation into equations for oxidation and reduction.

Oxidation reaction:

$\begin{matrix} 0 \\ Cu \end{matrix}\rightarrow \begin{matrix} +2+5-2 \\ { \left( Cu{ NO }_{ 3 } \right)  }_{ 2 }  \end{matrix}$

Reduction reaction:

$\begin{matrix} +1+5-2 \\ { HNO }_{ 3 } \end{matrix} \ \ \rightarrow \ \ \begin{matrix} +2-2 \\ NO \end{matrix}$

Now, balance the oxidation and reduction reaction equations by balancing all atoms except hydrogen and oxygen.

Oxidation: $\mathrm{Cu}+2 \mathrm{HNO}_{3} \rightarrow\left(\mathrm{CuNO}_{3}\right)_{2}$

Reduction: $\mathrm{HNO}_{3} \rightarrow \mathrm{NO}$

Now balance the oxygen atoms: If there are not equal oxygen atoms on both sides, add water molecules to balance the equation.

Oxidation: $\mathrm{Cu}+2 \mathrm{HNO}_{3} \rightarrow\left(\mathrm{CuNO}_{3}\right)_{2}$

Reduction: $\mathrm{HNO}_{3} \rightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}$

Balance the hydrogen atoms: If hydrogen atoms on both sides are not the same, balance the equation by adding protons or H^+.

Oxidation: $\mathrm{Cu}+2 \mathrm{HNO}_{3} \rightarrow\left(\mathrm{CuNO}_{3}\right)_{2}+2 \mathrm{H}^{+}$

Reduction: $\mathrm{HNO}_{3}+3 \mathrm{H}^{+} \rightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}$

Now balance the charge: To balance the charge, electrons are added to the side with more positive charge.

Oxidation: $\mathrm{Cu}+2 \mathrm{HNO}_{3} \rightarrow\left(\mathrm{CuNO}_{3}\right)_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$

Reduction: $\mathrm{HNO}_{3}+3 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}$

The electron gain must be made equal to electron lost. For this, the coefficients of all elements must be multiplied by the lowest common multiple.

Oxidation: $\left(\mathrm{Cu}+2 \mathrm{HNO}_{3} \rightarrow\left(\mathrm{CuNO}_{3}\right)_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\right) \times 3$

Reduction: $\left(\mathrm{HNO}_{3}+3 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}\right) \times 2$

So, we get:

Oxidation: $3 \mathrm{Cu}+6 \mathrm{HNO}_{3} \rightarrow 3\left(\mathrm{CuNO}_{3}\right)_{2}+6 \mathrm{H}^{+}+6 \mathrm{e}^{-}$

Reduction: $2 \mathrm{HNO}_{3}+6 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{NO}+4 \mathrm{H}_{2} \mathrm{O}$

Now add the oxidation and reduction equations, we get,

$3 \mathrm{Cu}+8 \mathrm{HNO}_{3}+6 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 3\left(\mathrm{CuNO}_{3}\right)_{2}+6 \mathrm{H}^{+}+6 \mathrm{e}^{-}+2 \mathrm{NO}+4 \mathrm{H}_{2} \mathrm{O}$

On further simplifying the equation, we get the balanced equation as:

$3 \mathrm{Cu}+8 \mathrm{HNO}_{3} \rightarrow 3\left(\mathrm{CuNO}_{3}\right)_{2}+2 \mathrm{NO}+4 \mathrm{H}_{2} \mathrm{O}$