Question

Balance the following redox reaction by any one method
 $ {C_2}{H_5}OH + CrO_7^{2 - }\xrightarrow{{{\text{acidic}}}}C{O_2} + C{r^{3 + }} $

Answer 1

Hint: The law of conservation of mass dictates that the quantity of each element should not change in a chemical reaction. Thus, each side of the chemical equation that is the product and the reactant side should represent the same quantity of a particular element. In a balanced equation charge on both sides of the reaction should also be conserved. Therefore, the same charge must be present on both sides of the reaction that is the same charge on the reactant side and the same charge on the product side.

Complete Step by step solution
We can balance a simple chemical equation by changing the scalar number for each chemical compound or element present in the reaction irrespective of the compound or element being present on the reactant or on the product side.
But for a redox reaction we will use another method in which we will do as follows,
First balance each half-equation for elements other than O and H.
Next balance each for oxygen using a water molecule that is $ {H_2}O $ .
Then balance each for H using positive hydrogen that is $ {H^ + } $ .
Finally, balance each for charge using electrons that is $ {e^ - } $ .
So, we apply the same steps on the equation given in question as follows,
 $ C{r_2}O_7^{2 - } \to 2C{r^{3 + }} $ and $ {C_2}{H_5}OH \to 2C{O_2} $
 $ C{r_2}O_7^{2 - } \to 2C{r^{3 + }} + 7{H_2}O $ and $ 3{H_2}O + {C_2}{H_5}OH \to 2C{O_2} $
 $ 14{H^ + } + C{r_2}O_7^{2 - } \to 2C{r^{3 + }} + 7{H_2}O $ and $ 3{H_2}O + {C_2}{H_5}OH \to 2C{O_2} + 12{H^ + } + 12{e^{ - 1}} $
 $ 6{e^{ - 1}} + 14{H^ + } + C{r_2}O_7^{ - 3} \to 2C{r^{3 + }} + 7{H_2}O $ and $ 3{H_2}O + {C_2}{H_5}OH \to 2C{O_2} + 12{H^ + } + 12{e^{ - 1}} $
The first equation is doubled to get 12 electrons in the reactant side so that both the halves have equal number of electrons, which can be nullified when the reactions are added.
So, we add the two halves of the equation and we get result as follows,
 $ 16{H^ + } + 2C{r_2}O_7^{2 - } + {C_2}{H_5}OH \to 4C{r^{3 + }} + 2C{O_2} + 11{H_2}O $ .

Note
A chemical equation consists of the chemical formula of the reactants that is the formulas of compounds and symbols of elements. The chemical equation also requires the chemical formulas of the compound and elements of the products. The reactants and the products are separated by arrow and the individual compound or element is divided by a plus sign.