Question

Balance the chemical equations by ion electron method.
(i). \[C{r_2}O_7^{2 - } + {C_2}{H_4}O + {H^ + } \to 2C{r^{3 + }} + {C_2}{H_4}{O_2} + {H_2}O\]
(ii). \[C{u_2}O + {H^ + } + NO_3^ - \to C{u^{2 + }} + NO + {H_2}O\]
(A) \[C{r_2}O_7^{2 - } + 3{C_2}{H_4}O + 8{H^ + } \to 2C{r^{3 + }} + 3{C_2}{H_4}{O_2} + 4{H_2}O\]
\[3C{u_2}O + 14{H^ + } + 2NO_3^ - \to 6C{u^{2 + }} + 2NO + 7{H_2}O\]
(B) \[C{r_2}O_7^{2 - } + 3{C_2}{H_4}O + 4{H^ + } \to 2C{r^{3 + }} + 3{C_2}{H_4}{O_2} + 2{H_2}O\]
\[3C{u_2}O + 14{H^ + } + 4NO_3^ - \to 6C{u^{2 + }} + 4NO + 7{H_2}O\]
(C) \[2C{r_2}O_7^{2 - } + 3{C_2}{H_4}O + 8{H^ + } \to 4C{r^{3 + }} + 3{C_2}{H_4}{O_2} + 4{H_2}O\]
\[6C{u_2}O + 14{H^ + } + 2NO_3^ - \to 6C{u^{2 + }} + 2NO + 7{H_2}O\]
(D) none of these

Answer 1

Hint: In order to obtain a balanced equation method, we have to solve the oxidizing half-cell and reducing half-cell separately and then add both the reactions, so that we can get a balanced equation.

Complete step by step answer:
We are going to balance the equation using the ion electron method. The ion electron will be based on the principle that the electron in the oxidation half-cell reaction is equal to the electron gained in the reduction half-cell reaction.

(i) Step 1:
Write down the chemical reaction.
\[C{r_2}O_7^{2 - } + {C_2}{H_4}O + {H^ + } \to 2C{r^{3 + }} + {C_2}{H_4}{O_2} + {H_2}O\]
Step 2:
Write the oxidation number of each element in the chemical reaction given.
Chromium will be reduced from the +6-oxidation state to +3 oxidation state. This means that this is the reduction half-cell. The Carbon from -1 oxidation state is oxidised to 0 oxidation state. This means that this reaction is an oxidation half-cell reaction.
Reduction half-cell
\[C{r_2}O_7^{2 - } \to 2C{r^{3 + }}\] (oxidation state of Cr from +6 to +3) .... (1)
Oxidation half-cell
\[{C_2}{H_4}O \to {C_2}{H_4}O_2^{}\] (oxidation state of Cr from -1 to 0) ...…. (2)
Step 3:
We can balance the charge of Cr in the equation (1) by adding six electrons in the LHS and further in order to balance the oxygen atom in the LHS, we are adding on the RHS. Then finally balance the Hydrogen atom on the RHS we are adding \[14{H^ + }\] ions on the LHS. Thus, the equation is written as
\[C{r_2}O_7^{2 - } + 6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O\] ……. (3)
We can balance the charge of the carbon atom in equation (2) by adding two electrons on the RHS. Further to balance the oxygen atom in the equation, we are adding a \[{H_2}O\]molecule on the LHS. Since there is more hydrogen on the LHS, we are adding \[2{H^ + }\]ions to the equation (2). Thus, the equation is written as
\[{C_2}{H_4}O + {H_2}O \to {C_2}{H_4}{O_2} + 2{e^ - } + 2{H^ + }\] ……. (4)
In order to balance the electrons in the equation (3) and (4), we have to multiply the equation (4) by 3, then equation (4) becomes
\[3{C_2}{H_4}O + 3{H_2}O \to 3{C_2}{H_4}{O_2} + 6{e^ - } + 6{H^ + }\]……... (5)
Adding equation (3) and (5) we get,
\[C{r_2}O_7^{2 - } + 6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O\]
\[3{C_2}{H_4}O + 3{H_2}O \to 3{C_2}{H_4}{O_2} + 6{e^ - } + 6{H^ + }\]
We are getting the final equation as
\[C{r_2}O_7^{2 - } + 3{C_2}{H_4}O + 8{H^ + } \to 2C{r^{3 + }} + 3{C_2}{H_4}{O_2} + 4{H_2}O\]
Similarly, now take the other equation given in the question
\[C{u_2}O + {H^ + } + NO_3^ - \to C{u^{2 + }} + NO + {H_2}O\]

(ii) Step 1: Write down the chemical reaction.
\[C{u_2}O + {H^ + } + NO_3^ - \to C{u^{2 + }} + NO + {H_2}O\]
Step 2:
Write the oxidation number of each element in the chemical reaction given.
In the chemical reaction Cu from 0 state is oxidized to +2 oxidation state. This oxidation half-cell and Nitrogen from +5 is reduced to +2 state. This is the reduction half-cell.
Oxidation half-cell.
\[C{u_2}O \to C{u^{2 + }}\]……... (6)
Reduction half cell
\[NO_3^ - \to NO\]……. (7)
First, we have to balance the equation other than Oxygen and Hydrogen, therefore the equation (6) becomes,
\[C{u_2}O \to 2C{u^{2 + }}\]…… (8)
Step 3:
Now we can balance the oxygen and hydrogen atom, we can add \[{H_2}O\]in the RHS, to balance the oxygen in the LHS. Now add \[{H^ + }\]to balance the hydrogen atom in the RHS. Now we can balance the charge of the equation. In order to balance the charge on both sides add 2 electrons in the RHS. Thus, the equation is written as
\[C{u_2}O + 2{H^ + } \to 2C{u^{2 + }} + {H_2}O + 2{e^ - }\] …… (9)
 Now balance the oxygen atom in the equation and add two \[{H_2}O\] on the RHS. Now to balance the hydrogen, we can add \[4{H^ + }\]on the LHS. In order to balance the charge on the nitrogen atom we can add 3 electrons on the LHS in the equation (7). Thus, the equation is given as
\[NO_3^ - + 4{H^ + } + 3{e^ - } \to NO + 2{H_2}O\] …... (10)
Now to balance the electrons on both equation (9) and (10), we can multiply equation (9) by 3 and equation (10) by 2. therefore, we get
\[3C{u_2}O + 6{H^ + } \to 6C{u^{2 + }} + 3{H_2}O + 6{e^ - }\]………. (11)
\[2NO_3^ - + 8{H^ + } + 6{e^ - } \to 2NO + 4{H_2}O\]……. (12)
Now adding equation (11) and (12), we get
\[3C{u_2}O + 14{H^ + } + 2NO_3^ - \to 6C{u^{2 + }} + 2NO + 7{H_2}O\]

Therefore, the correct answer is option (A)
(i) \[C{r_2}O_7^{2 - } + 3{C_2}{H_4}O + 8{H^ + } \to 2C{r^{3 + }} + 3{C_2}{H_4}{O_2} + 4{H_2}O\]
(ii) \[3C{u_2}O + 14{H^ + } + 2NO_3^ - \to 6C{u^{2 + }} + 2NO + 7{H_2}O\].

Note: While balancing the equation always balances the charge, we should not only balance the atoms present in the equation, the charge is also an important factor. Not balancing it can lead to an unbalanced equation and hence your answer would not be correct.