Complete and balance the following chemical equations:
(a) Fe²⁺ + MnO₄⁻ + H⁺ →
(b) MnO₄⁻ + H₂O + I⁻ →
Assign the reason for the following :
Copper (I) ion is not known in aqueous solution.
Complete the following chemical equations :
(i) MnO₄⁻(aq) + S₂O₃²⁻(aq) + H₂O(l) →
(ii) Cr₂O₇²⁻(aq) + Fe²⁺(aq) + H⁺(aq) →
Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number.
Write the formula of an oxo-anion of Chromium (Cr) in which it shows the oxidation state equal to its group number.
Complete the following reactions—
(i) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ →
(ii) 2CrO₄²⁻ + 2H⁺ →
(iii) 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ →
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with (i) an iodide (ii) H₂S.
Give reasons :
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Transition metals show variable oxidation states.
(iii) Actinoids show irregularities in their electronic configurations.
(i) Give reasons for the following :
(a) Compounds of transition elements are generally coloured.
(b) MnO is basic while Mn₂O₇ is acidic.
(ii) Calculate the magnetic moment of a divalent ion in aqueous medium if its atomic number is 26.
Complete the following reactions—
(i) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ →
(ii) 2CrO₄²⁻ + 2H⁺ →
(iii) 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ →
Write the preparation of following :
(i) KMnO₄ from K₂MnO₄
(ii) Na₂CrO₄ from FeCr₂O₄
(iii) Cr₂O₇²⁻ from CrO₄²⁻
Consider the reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -> 2Cr³⁺ + 7H₂O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O₇²⁻?
Calculate emf of the following cell
Cd/\(Cd^{2+}\) (.10 M)//\(H_+\) (.20 M)/\(H_2\) (0.5 atm)/Pt
[Given E° for \(Cd^{2+}\) /Cd = -0.403V]
(i) Complete the following equations :
(a) 2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ →
(b) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ →
(ii) Based on the data, arrange Fe²⁺, Mn²⁺ and Cr²⁺ in the increasing order of stability of +2 oxidation state.
E°(Cr³⁺/Cr²⁺) = -0.4 V
E°(Mn³⁺/Mn²⁺) = +1.5 V
E°(Fe³⁺/Fe²⁺) = + 0.8 V