Question

Which of the following is a redox reaction:
A.$2Cr{O_4}^{2 - } + 2{H^ + } \to C{r_2}{O_7}^{2 - } + {H_2}O$
B.$CuS{O_4} + 4N{H_3} \to \left[ {Cu{{(N{H_3})}_4}} \right]S{O_4}$
C.$2N{a_2}{S_2}{O_3} + {I_2} \to N{a_2}{S_4}{O_6} + 2NaI$
D.$C{r_2}{O_7}^{2 - } + 2OH \to 2Cr{O_4}^{2 - } + {H_2}O$

Answer 1

Hint: We have to remember that a redox reaction is a reaction in which oxidation and reduction takes place simultaneously. In a redox chemical reaction, the change in the oxidation states of the reactants is used to identify a redox reaction. The substance getting reduced is known as an oxidizing agent and the substance getting oxidized is known as the reducing agent.

Complete step by step answer:
As we know that a redox reaction is characterized by transfer of electrons between the reactants. This transfer can be detected by observing the oxidation states of the reactants. Oxidation depicts loss of electrons and hence the oxidation state increases. Reduction depicts gain of electrons hence oxidation state decreases. The presence of both in a reaction is used to identify a redox reaction.
Let us evaluate each of the given reaction:
 $2Cr{O_4}^{2 - } + 2{H^ + } \to C{r_2}{O_7}^{2 - } + {H_2}O$:
Now we can calculate the oxidation state of Cr in $Cr{O_4}^{2 - }$ :
Let x be the oxidation state of Cr.
The oxidation state of O is -2
Therefore,
\[x + 4\left( { - 2} \right) = - 2\]
On simplification we get,
$x = + 6$
Oxidation state of Cr in $C{r_2}{O_7}^{2 - }$ : Let x be the oxidation state of Cr.
Therefore,
$2x + 7( - 2) = - 2$
On simplification we get,
$x = + 6$
There is no change in the oxidation number of Cr hence it is not a redox reaction. Therefore, the option A is incorrect.
 $CuS{O_4} + 4N{H_3} \to \left[ {Cu{{(N{H_3})}_4}} \right]S{O_4}$
Now we can calculate the oxidation state of $Cu$in $CuS{O_4}$ :
The sulphate ion has a -2 oxidation state and hence to neutralize the ion, Cu has an oxidation state of +2.
Now we can calculate the oxidation state of Cu in $\left[ {Cu{{(N{H_3})}_4}} \right]S{O_4}$ : The sulphate ion has a -2 oxidation state and $N{H_3}$ has 0 oxidation state. Hence Cu has an oxidation state of +2.
 There is no change in the oxidation number of Cr hence it is not a redox reaction. Therefore, the option B is incorrect.
 $2N{a_2}{S_2}{O_3} + {I_2} \to N{a_2}{S_4}{O_6} + 2NaI$
The oxidation state of Na2 has an oxidation state of +2 before and after the reaction.
Oxidation state of S in $N{a_2}{S_2}{O_3}$: Sulphur bonded to three oxygen is considered to have +6 and other sulphur has -2.
Oxidation state of S in$N{a_2}{S_4}{O_6}$: Let x be the oxidation state of x.
$2( + 1) + 4x + 6( - 2) = 0$
$4x = 10$
$x = + 2.5$
The oxidation state of S has increased hence this is the oxidation reaction.
The oxidation state of ${I_2} = 0$ that of $I$ in $NaI$ Is -1. The oxidation state has decreased hence this is the reduction reaction.
Hence this is a redox reaction. Therefore, the option C is correct.
 $C{r_2}{O_7}^{2 - } + 2OH \to 2Cr{O_4}^{2 - } + {H_2}O$
Oxidation state of Cr in $C{r_2}{O_7}^{2 - }$ : Let x be the oxidation state of Cr.
$2x + 7( - 2) = - 2$
$x = + 6$
Oxidation state of Cr in $Cr{O_4}^{2 - }$ : Let x be the oxidation state of Cr.
The oxidation state of O is -2.
Therefore,
\[x + 4\left( { - 2} \right) = - 2\]
On simplification we get,
$x = + 6$
There is no change in the oxidation number of Cr hence it is not a redox reaction. Therefore, the option D is not correct.
Hence, the correct option is option (C).

Note:
We must be noted that there are some general rules to be followed when calculating the oxidation number of any element. Two important of them are,
(i) In a neutral compound, all oxidation numbers must add up to zero. A neutral compound does not have a plus or minus charge.
(ii)In an ion, all the oxidation numbers must add up to the charge of the ion.