Question

Which of the following are common oxidizing agents used in redox titrations?
A.${{K}_{2}}C{{r}_{2}}{{O}_{7}},KMn{{O}_{4}}$, iodine
B.$FeS{{O}_{4}},KMn{{O}_{4}}$, sodium thiosulphate
C. oxalic acid, $KMn{{O}_{4}},CuS{{O}_{4}}$
D. Mohr’s salt,$KI$, sodium sulphate

Answer 1

Oxidizing agent is the species that gains electrons and is capable to give a reduction reaction. The reactions in which reduction and oxidation occur simultaneously are called redox reactions. The redox reactions can be identified by the reducing and oxidizing capability of compounds by the change in their oxidation states.

Complete answer:
Titration is a method used for analyzing the concentration of various solutions. A redox titration employs a redox reaction, where oxidation – reduction both happens. We have been given to find the oxidizing agent used in the redox titration.
An oxidizing agent is the species that gets reduced by gaining electrons and has a decrease in its oxidation state. Therefore, compounds having elements in their highest oxidation number, will be the preferred oxidizing agents used for the redox titrations.
We will calculate the oxidation states of the central atoms by the fact that a molecule is neutral so the sum of the oxidation states of all atoms is equal to zero. If they are in their highest state then they are suitable oxidizing agents.
-${{K}_{2}}C{{r}_{2}}{{O}_{7}}$, has K in +1, O in -2 state, for Cr, we have 2 (+1) + 2(Cr) + 7 (-2) = 0
2Cr + 2 – 14 = 0,
2Cr = 12
Cr = +6, which is its maximum state.
-$KMn{{O}_{4}}$ has K in +1, O in -2 state, so, Mn will have 1 + Mn + 4 (-2) = 0
Mn + 1 – 8 = 0
Mn = +7, which is its maximum oxidation state.
-Iodine is in 0 oxidation state, as it is a monatomic molecule, so the oxidation states will cancel out and become 0. But it is electronegative and can gain electrons so it is an oxidizing agent.
-$FeS{{O}_{4}}$, has sulphate ion in -2 oxidation state, so Fe + (-2) = 0
Fe = +2, this is not the highest oxidation state of Fe.
-$KMn{{O}_{4}}$ has Mn in +7 (highest state)
-Sodium thiosulphate $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$, has Na in +1 and O in -2 state, so, 2 (1) + 2 (S) + 3 (-2) = 0
2 S + 2 – 6 = 0
2 S = 4
S = +2, this is not the highest oxidation state of Sulphur.
-Oxalic acid ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ has H in +1, O in -2, so 2 + 2 C + (-8) = 0
2 C = 6
C = +3, this is not the highest oxidation state of carbon.
-$KMn{{O}_{4}}$ has Mn in +7 (highest state)
-$CuS{{O}_{4}}$ has sulphate in -2 state, so Cu has +2, which is its highest oxidation state.
-Mohr’s salt ${{(N{{H}_{4}})}_{2}}Fe{{(S{{O}_{4}})}_{2}}$ has ammonium in +1 and sulphate in -2 state, so, Fe +2 – 4 = 0
Fe = +2, this is not its highest oxidation state.
-KI is potassium iodide which is not an oxidizing agent.
-Sodium sulphate$N{{a}_{2}}S{{O}_{4}}$has Na in +1, O in -2, so, 2 + S + (-8) = 0
S = +6.
Hence, .${{K}_{2}}C{{r}_{2}}{{O}_{7}},KMn{{O}_{4}}$, iodine are common oxidizing agents used in redox titrations.

Note:
The highest oxidation state for Fe is +3, for sulphur is +6, for carbon is +4. Oxidation state is the ability to gain or lose electrons. A redox reaction involves reduction which decreases the oxidation number so a compound with a higher oxidation number will act as a good oxidizing agent.