Question 1

500 mL of 0.2 M MnO_4^- solution in basic medium when mixed with 500 mL of 1.5 M KI solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard

Accepted Answer

First, we calculate the millimoles of the reactants: Millimoles of MnO _4^- = 500 mL 0.2 M = 100 mmol Millimoles of I ^- = 500 mL 1.5 M = 750 mmol In a basic medium, MnO _4^- is reduced to MnO _2. The change in oxidation state of Mn is from +7 to +4, so its n-factor is 3. The problem explicitly states that iodide ions are oxidized to molecular iodine ( I _2). The change in oxidation state for iodine is from -1 to 0, so the n-factor for I ^- is 1. Equivalents of MnO _4^- = 100 3 = 300 meq Equival

Question 2

Given below are two statements: Statement I: Sodium dichromate and potassium dichromate are classified as primary standards in titrimetric analysis. Statement II: Phenolphthalein is a weak base, there

Accepted Answer

Statement I is false because sodium dichromate is deliquescent (absorbs moisture from the atmosphere) and cannot be weighed accurately, so it is not used as a primary standard. Potassium dichromate, however, is a primary standard. Statement II is false because phenolphthalein is a weak organic acid, not a weak base. It remains largely undissociated in an acidic medium (colorless) and dissociates in a basic medium to give a pink color. Therefore, both Statement I and Statement II are false. Answe

Question 3

In order to oxidise a mixture of 1 mole each of FeC_2O_4, Fe_2(C_2O_4)_3, FeSO_4 and Fe_2(SO_4)_3 in acidic medium, the number of moles of KMnO_4 required is

Accepted Answer

In acidic medium, KMnO_4 acts as an oxidising agent and gets reduced to Mn^ 2+ . The n-factor for KMnO_4 is 5. The n-factors for the given reducing agents are calculated based on the total number of moles of electrons lost per mole of the compound: For FeC_2O_4: Fe^ 2+ Fe^ 3+ + e^ - C_2O_4^ 2- 2CO_2 + 2e^ - Total electrons lost = 3, so n-factor = 3. Equivalents = 1 3 = 3. For Fe_2(C_2O_4)_3: Fe^ 3+ is not oxidised. 3C_2O_4^ 2- 6CO_2 + 6e^ - Total electrons lost = 6, so n-factor = 6. Equivalents

Question 4

500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO _ 4 solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na _ 2 ~S _ 2 O _ 3 solution in the presence of starc

Accepted Answer

MnO_4^- + I^- MnO_2 + I_2 I_2 + S_2O_3^ 2- S_4O_6^ 2- + I^- gram eq of KMnO_4 = gram eq of Na_2S_2O_3 0.2 500 1000 3 = 0.1 V 1 V = 3 L

Question 5

One mole of Cl _ 2 ( ~g ) was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of Cl ^ - , ClO ^ - and OH ^ - are respectively (assume volume remains constant)

Accepted Answer

Cl₂ undergoes disproportionation in cold KOH: Cl_2 + 2OH^- Cl^- + ClO^- + H_2O Initial: 1 mole Cl₂, 2 L × 2 M KOH = 4 moles OH⁻ Molar ratio Cl₂:OH⁻ = 1:2 (stoichiometric) Products formed: 1 mole Cl⁻, 1 mole ClO⁻; Excess OH⁻ = 4 - 2 = 2 moles In 2 L solution: [Cl⁻] = 0.5 M, [ClO⁻] = 0.5 M, [OH⁻] = 1 M

Question 6

X and Y are the number of electrons involved, respectively during the oxidation of I ^ - to I _ 2 and S ^ 2- to S by acidified K _ 2 Cr _ 2 O _ 7 . The value of X + Y is \_\_\_\_.

Accepted Answer

In acidic medium, K_2Cr_2O_7 acts as a strong oxidizing agent where Cr_2O_7^ 2- is reduced to Cr^ 3+ . The reduction half-reaction is: Cr_2O_7^ 2- + 14H^+ + 6e^- 2Cr^ 3+ + 7H_2O. For the oxidation of I^- to I_2: The balanced oxidation half-reaction is 2I^- I_2 + 2e^-. To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3: 6I^- 3I_2 + 6e^-. Thus, the number of electrons involved in the balanced redox reaction for I^- oxidation is X =

Question 7

200 cc of x 10^ -3 M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here x= \_\_\_\_.

Accepted Answer

Mohr's salt (FeSO_4 (NH_4)_2SO_4 6H_2O) is a source of Fe^ 2+ ions. The redox reaction with potassium dichromate in acidic medium is: 6Fe^ 2+ + Cr_2O_7^ 2- + 14H^+ 6Fe^ 3+ + 2Cr^ 3+ + 7H_2O. Moles of Fe^ 2+ from Mohr's salt solution: n(Fe^ 2+ ) = 0.6 M 0.750 L = 0.45 mol . From the stoichiometry, 6 moles of Fe^ 2+ require 1 mole of Cr_2O_7^ 2- . Moles of Cr_2O_7^ 2- required: n(Cr_2O_7^ 2- ) = 0.45 6 = 0.075 mol . Molarity of potassium dichromate: M = 0.075 mol 0.200 L = 0.375 M = 375 10^ -3 M .

Question 8

Identify the correct statements : A. Hydrated salts can be used as primary standard. B. Primary standard should not undergo any reaction with air. C. Reactions of primary standard with another substan

Accepted Answer

Primary standard must be soluble for standard solution formation

Redox Reactions — Previous Year Questions & Solutions