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Question: Define auto redox reaction. Balance the following equation by ion- electron or oxidation number meth...

Define auto redox reaction. Balance the following equation by ion- electron or oxidation number method?
K2Cr2O7(aq)+FeSO4(aq)+H2SO4(aq)→K2SO4(aq)+Cr2(SO4)3(s)+Fe2(SO4)3(s)+H2O(l){K_2}C{r_2}{O_7}(aq) + FeS{O_4}(aq) + {H_2}S{O_4}(aq) \to {K_2}S{O_4}(aq) + C{r_2}{(S{O_4})_3}(s) + F{e_2}{(S{O_4})_3}(s) + {H_2}O(l)

Explanation

Solution

In order to answer this question, this auto redox reaction is related to the redox reaction i.e. reduction and oxidation in a reaction. Also, the balancing of chemical reactions is based on the oxidation number of the atoms on both sides of the reactions.

Complete answer:
Let’s understand the complete solution in detail.
First is about the auto redox reaction. An auto redox reaction is a redox reaction in which a substance acts as both the oxidizing agent and the reducing agent.
Now, the second is to balance the given chemical equation according to the oxidation number method.
K2Cr2O7(aq)+FeSO4(aq)+H2SO4(aq)→K2SO4(aq)+Cr2(SO4)3(s)+Fe2(SO4)3(s)+H2O(l){K_2}C{r_2}{O_7}(aq) + FeS{O_4}(aq) + {H_2}S{O_4}(aq) \to {K_2}S{O_4}(aq) + C{r_2}{(S{O_4})_3}(s) + F{e_2}{(S{O_4})_3}(s) + {H_2}O(l)
First, we need to identify the oxidizing and reducing agent.
We observe that chlorine atoms get reduced as its oxidation number of this elements decreases from +6+ 6 in K2Cr2O7{K_2}C{r_2}{O_7} to +3+ 3 in Cr2(SO4)3C{r_2}{(S{O_4})_3} . Therefore, chromate ions act as oxidizing agents and undergo the following reactions.
Oxidation half reaction-
Cr2O7(aq)+6e−+14H+(aq)→2Cr3+(aq)+7H2O(l)C{r_2}{O_7}(aq) + 6{e^ - } + 14{H^ + }(aq) \to 2C{r^{3 + }}(aq) + 7{H_2}O(l)
Now, add water to the product side and protons on the reactant side to balance the number of oxygen atoms.
Now, the Iron undergoes oxidation and acts as a reducing agent.
Reduction half reaction-
Fe2+(aq)→Fe3+(aq)+e−F{e^{2 + }}(aq) \to F{e^{3 + }}(aq) + {e^ - }
Electrons should cancel out in the net reactions. Electrons will cancel out if the coefficient of the oxidation half reaction is expanded by a factor.
6Fe2+(aq)→6Fe3+(aq)+6e−6F{e^{2 + }}(aq) \to 6F{e^{3 + }}(aq) + 6{e^ - }
Combining oxidation and reduction half reaction, we get
Cr2O7(aq)+6e−+14H+(aq)+6Fe2+(aq)→2Cr3+(aq)+7H2O(l)+6Fe3+(aq)+6e−C{r_2}{O_7}(aq) + 6{e^ - } + 14{H^ + }(aq) + 6F{e^{2 + }}(aq) \to 2C{r^{3 + }}(aq) + 7{H_2}O(l) + 6F{e^{3 + }}(aq) + 6{e^ - }
Electrons get cancel out on both the sides,
Cr2O7(aq)+14H+(aq)+6Fe2+(aq)→2Cr3+(aq)+7H2O(l)+6Fe3+(aq)C{r_2}{O_7}(aq) + 14{H^ + }(aq) + 6F{e^{2 + }}(aq) \to 2C{r^{3 + }}(aq) + 7{H_2}O(l) + 6F{e^{3 + }}(aq)
Now, pair the ions with the opposite of respective charges.
Anion reactants will pair with opposite charges K+(aq){K^ + }(aq) whereas cation reactants pair with sulfate ions SO42−(aq)SO_4^{2 - }(aq) .
Hence, the net chemical reaction will be
K2Cr2O7(aq)+6FeSO4(aq)+7H2SO4(aq)→K2SO4(aq)+Cr2(SO4)3(s)+3Fe2(SO4)3(s)+7H2O(l){K_2}C{r_2}{O_7}(aq) + 6FeS{O_4}(aq) + 7{H_2}S{O_4}(aq) \to {K_2}S{O_4}(aq) + C{r_2}{(S{O_4})_3}(s) + 3F{e_2}{(S{O_4})_3}(s) + 7{H_2}O(l) .

Note:
It must be remembered that a disproportionation reaction would follow the definition of auto-redox reactions in which the same species get oxidized and reduced. Observe the change in oxidation number carefully on both sides of the reactions.