8

Q1. The number of moles of KMnO₄ reduced by one mole of KI in alkaline medium is

• one
• two
• five
• one fifth

Solution

The number of moles of KMnO₄ reduced by one mole of KI in alkaline medium is one fifth.

Q2. The conversion of K₂Cr₂O₇ into Cr₂ (SO₄)₃ is a process of

• Oxidation
• Reduction
• Decomposition
• Substitution

Solution

Oxidation number of Cr changes from +6 in  K₂Cr₂O₇ to +3 in Cr₂ (SO₄)₃. So, this is an example of reduction.

Q3. By what unit the oxidation number changes for underline elements in the following reaction- Zn + HNO₃ → NH₄NO₃ + Zn (NO₃)₂ + H₂O

• 4 units
• 2 units
• 6 units
• 8 units

Solution

In HNO₃ the oxidation state of N is +5 and the oxidation number of N is -3 in NH₄NO₃ hence there is change in oxidation number by 8 units.

Q4. The oxidation half reaction for following reaction is   Fe²⁺(aq) + Cr₂O₇^2-(aq) → Fe³⁺ (aq) + Cr³⁺(aq)

• Fe²⁺ (aq) → Fe³⁺(aq)
• Cr₂O₇^2-(aq) → Cr³⁺(aq)
• Fe³⁺(aq) → Fe²⁺ (aq)
• Cr³⁺(aq) → Cr₂O₇^2-(aq)

Solution

Since in the given reaction Fe²⁺ is oxidized to Fe³⁺ . Hence the oxidation half reaction can be represented as Fe²⁺ (aq) → Fe³⁺(aq)

Q5. Balanced ionic equation for the reaction of Potassium Dichromate (VI) K₂Cr₂O₇ with Sodium Sulphite Na₂SO₃ in acid solution to give Chromium (III) ion and sulphate ion is

• Cr₂O₇^-2_(aq) + 3SO₃^2-_(aq) + 8H⁺_(aq) → 2Cr³⁺_(aq) + 3SO₄^2-_(aq) + 4H₂O_(liq)
• Cr₂O₇^-2_(aq) + 2SO₃^2-_(aq) + 8H⁺_(aq) → 2Cr³⁺_(aq) + 3SO₄^2-_(aq) + 4H₂O_(liq)
• Cr₂O₇^-2_(aq) + 3SO₃^2-_(aq) + 5H⁺_(aq) → 2Cr³⁺_(aq) + 3SO₄^2-_(aq) + 4H₂O_(liq)
• Cr₂O₇^-2_(aq) + 3SO₃^2-_(aq) + 8H⁺_(aq) → 2Cr³⁺_(aq) + 5SO₄^2-_(aq) + 4H₂O_(liq)

Solution

Cr₂O₇^-2_(aq) + 3SO₃^2-_(aq) + 8H⁺_(aq) → 2Cr³⁺_(aq) + 3SO₄^2-_(aq) + 4H₂O_(liq)

Q6. In Ni(CO)₄ the oxidation number of Ni is

• 4
• 2
• 8
• 0

Solution

The charge on CO group is 0. So, x + 0 = 0; x = 0

Q7. If an element is in its lowest oxidation state under proper conditions, it can act as

• Reducing agent
• Oxidizing agent
• Both the reducing and oxidizing agent
• Neither oxidizing nor reducing agent

Solution

Since the element is in lowest oxidation state, it can further oxidize into higher oxidation state hence it can act as reducing agent.

Q8. The oxidation number of Mg in Mg⁺² is

• +2
• +1
• -2
• -1

Solution

For ions composed of only one atom, the oxidation number is equal to the charge on the ion.e.g. oxidation number of Mg in Mg⁺² is +2.

Q9. The oxidation and reduction half reactions of the reaction 2Al_(s) + 3Cu²⁺_(aq) → 2Al³⁺_(aq) + 3Cu_(s) are

• Oxidation: 2Al_(s)  →   Al³⁺_(aq) + 3e^- ] x 2    Reduction: Cu²⁺ + 2e^- → Cu_(s) ] x 3          
• Oxidation: Al_(s)   →  Al³⁺ + 3e^- ] x 2           Reduction: Cu²⁺ + 2e^- → Cu_(s) ] x 3          
• Oxidation: Al_(s) → Al³⁺_(aq) + 3e^- ] x 2            Reduction: 2Cu²⁺_(aq) + 2e^-  →  Cu_(s) ] x 3      
• Oxidation: Al_(s)  →    Al³⁺_(aq) + 3e^- ] x 2         Reduction: Cu²⁺_(aq) + 2e^- →  Cu_(s) ] x 2        

Solution

Oxidation: Al_(s) → Al³⁺ + 3e^- ] x 2               Reduction: Cu²⁺ + 2e^-  → Cu_(s) ] x 3          

Q10. The algebraic sum of the oxidation number of all the atoms in a compound must be

• 1
• -2
• -1/2
• 0

Solution

The algebraic sum of the oxidation number of all the atoms in a compound must be zero.

Q11. In an electrochemical cell represented as, Zn | Zn²⁺₍c₁₎ || Cu²⁺₍c₂₎ | Cu so in this cell,

• Zn is oxidized
• Zn is reduced
• Cu is oxidized
• Cu is reduced

Solution

Zn is oxidized

Q12. In MnO₄^-4 the oxidation number of Mn is-

• 4
• 7
• 2
• 8

Solution

x + 4(-2) = -4 x = 4

Q13. We can balance the following ionic equation by adding   6Fe²⁺(aq) + Cr₂O₇^2-(aq)  → 6 Fe³⁺(aq) +2Cr ³⁺(aq)

• 7 H₂O on RHS and 14 H⁺ on LHS
•  7 H₂O on LHS and 14 H⁺ on RHS
• 3H₂O on RHS and 6 H⁺ on LHS
• 3H₂O on LHS and 6 H⁺ on RHS

Solution

For balancing the ionic reaction 14 H⁺ should be added on left hand side of equation and 7H₂O on right hand side of equation.

Q14. A reaction in which same element is simultaneously oxidized as well as reduced is called

• Combination Reaction
• Decomposition Reaction
• Disproportionation Reaction
• Displacement

Solution

In disproportionation reaction as the same element have different oxidation number so same element get oxidized as well as reduced.

Q15. In the reaction H₂S_(g) + Cl_2(g) → 2HCl_(g) + S_(s), H₂S is

• Oxidised
• Reduced
• None of them
• both a and b

Solution

H₂S is oxidised as there is removal of hydrogen.

Q16. The oxidation number of Hydrogen is +1 except in

• Metal hydride
• Non-metal hydride
• Metalloid hydride
• Hydrogen bonded compound

Solution

The oxidation number of Hydrogen is +1 except in binary metal hydrides. for example NaH, here the oxidation state of H is -1 as metals are more electropositive than hydrogen.

Q17. A substance (atom, ion or molecule) which can readily loose electrons to other substances is called

• Oxidising agent
• Reducing agent
• None of them
• both a and b

Solution

Reducing agent is donor of electrons.

Q18. Which one of the following is reducing agent?

• Ozone
• Chlorine
• FeCl₃
• Na₂S₂O₃

Solution

Na₂S₂O₃ acts as reducing agent because S₂O₃^2- ion oxidizes into S₄O₆^2- ion.

Q19. How many electrons are transferred during following change?2I^-_(aq) → I_{2 (s)}

• 4 electrons
• 0 electrons
• 2 electrons
• 1 electron

Solution

In this half reaction 2 electrons are transferred and oxidation number of iodine changes from -1 to 0.

Q20. For the reaction Fe²⁺ _(aq) + Cr₂O₇^2- _(aq) → Fe³⁺ _(aq) + Cr³⁺ _(aq) , the reduction half reaction is,

• Cr₂O₇^2- _(aq) → Cr³⁺ _(aq)
• Fe²⁺ _(aq) → Fe³⁺ _(aq)
• Fe²⁺ _(aq) + Cr₂O₇^2- _(aq) → Fe³⁺ _(aq) + Cr³⁺ _(aq)
• None of these

Solution

For the reaction Fe²⁺ _(aq) + Cr₂O₇^2- _(aq) → Fe³⁺ _(aq) + Cr³⁺ _(aq) , the reduction half reaction is Cr₂O₇^2- _(aq) → Cr³⁺ _(aq)

Q21. When Zinc rod is dipped in copper nitrate solution, the blue colour of copper nitrate solution starts fading. The substance that undergoes oxidation is

• Zn_(s)
• Zn²⁺
• Cu_(s)
• Cu²⁺

Solution

Zn_(s) undergoes oxidation because there is release of electrons.

Q22. Which of the following is NOT a redox reaction?

• Burning of candle
• Rusting of iron
• Dissolving a salt in water
• Dissolving zinc in dil. H₂SO₄

Solution

Dissolving a salt in water is not a redox reaction because no oxidation or reduction takes place either of water or salt.

Q23. In which of the following method, two half equations are balanced separately and then added together to give balanced equation?

• Half reaction method
• Reducing agent method
• Oxidizing agent method
• Reluctant method

Solution

In half reaction method, the two half equations are balanced separately and then added together to give balanced equation.

Q24. To balance H in Cr₂O₇^2- _(aq) → Cr³⁺ _(aq) + 7H₂O (in acidic medium), we will

• Add 14H⁺ on left side
• Add 4H⁺ on left side
• Add 14H⁺ on right side
• Add 4H⁺ on right side

Solution

To balance H in Cr₂O₇^2- _(aq) → Cr³⁺ _(aq) + 7H₂O (in acidic medium), we will add 14H⁺ on left side. Cr₂O₇^2- _(aq) + 14H⁺ → Cr³⁺ _(aq) + 7H₂O

Q25. The process in which the strength of an unknown solution is calculated using a known standard solution is known as

• titration
• oxidation
• reduction
• none of these

Solution

The process in which the strength of an unknown solution is calculated using a known standard solution is known as titration.

Q26. Two half cell are Al³⁺_(aq) | Al and Mg²⁺_(aq) | Mg. The reduction potential of these half cells are -1.66V and -2.36V. The EMF of the cell is:

• +0.70V
• -0.70V
• +1.70V
• -1.70V

Solution

The EMF of the cell is, E_cell= E_cathode - E_anode So, E_cell = -1.66 - (-2.36) = 2.36-1.66 = +0.70V.

Q27. Given the standard reduction potentials: Zn²⁺/Zn = - 0.74 V, Cl₂/Cl^- = 1.36 V, H⁺/1/2H₂ = 0 V and Fe³⁺/Fe²⁺ = 0.77 V. The order of increasing strength as reducing agent is

• Zn, H₂, Fe²⁺, Cl^-
• H₂, Zn, Fe²⁺, Cl^-
• Cl^-, Fe²⁺,Zn, H₂
• Cl^-, Fe²⁺, H_2, Zn

Solution

Correct order is: Cl^-, Fe²⁺, H_2, Zn

Q28. 2 K₄ [Fe (CN)₆]_(aq) + H₂O_{2 (aq)} → 2 K₃ [Fe (CN)₆]_(aq) + 2KOH_(aq) is an example of

• Oxidation
• Reduction
• Replacement
• Addition

Solution

2 K₄ [Fe (CN)₆]_(aq) + H₂O_{2 (aq)} → 2 K₃ [Fe (CN)₆]_(aq) + 2KOH_(aq) is an example of  oxidation because "oxidation" is defined as the addition of removal of electropositive element from a substance.There is removal of electropositive element potassium from Potassium ferrocyanide.

Q29. In Daniel cell, oxidation takes place at

• Only at anode
• Only at cathode
• On both anode and cathode
• It depends on the salts and their solutions

Solution

In Daniel cell, oxidation takes place at anode.

Q30. Calculate E^o for the cell, Al | Al³⁺_(1M) || Cu²⁺_(1M) | Cu. Given E^o_Al³⁺/Al and E^o_Cu²⁺/Cu as -1.66 V and 0.34 V respectively.

• -2.00 V
• +2.00 V
• +1.00 V
• -1.00 V

Solution

E^o_cell = E^o_Right - E^o_Left = 0.34-(-1.66) = +2.00 v

Q31. To balance the oxygen atom in the given reaction in acidic medium, Cr₂O₇^2- _(aq) → Cr³⁺ _(aq) we will,

• Add O on left side
• Add water (H₂O) on left side
• Add water (H₂O) on right side
• Add O on right side

Solution

To balance the oxygen atom in the given reaction in acidic medium, Cr₂O₇^2- _(aq) → Cr³⁺ (aq) + H₂O

Q32. To balance the charge in the oxidation half reaction Fe²⁺ _(aq) → Fe³⁺_(aq)

• Add an electron on reactant side
• Add an electron on product side
• Add two electrons on reactant side
• Add two electrons on product side

Solution

Fe²⁺ _(aq) → Fe³⁺_(aq) + e^- One electron is added on product side.

Q33. In acid solution, the reactionMnO₄^- → Mn²⁺ shows:

• Oxidation by 5 electrons
• Oxidation by 3 electrons
• Oxidation by 7 electrons
• Reduction by 5 electrons

Solution

Since there is reduction of Mn from +7 to +2 state there is reduction by 5 electrons.

Q34. Oxidation number of C in CO₂ is:

• +2
• +4
• -2
• -4

Solution

Oxygen is assigned the oxidation number -2 in its simple oxides. so, x + 2(-2)=0 x = 4

Q35. The nature of electolytes used in a salt bridge are:

• Non electrolytes
• Inert electrolytes
• Weak electrolytes
• Any electrolyte can be used

Solution

Inert electrolytes are used as their ions are not involved in any electrochemical change.

Q36. Pick the odd one out

• Combination redox reaction
• Displacement redox reaction
• Catalysis redox reaction
• Decomposition redox reaction

Solution

All are the types of redox reactions except catalysis redox reaction.

Q37. The standard electrode potentials, K⁺/K= -2.93V, Ag⁺/Ag = 0.80V, the electrode which is negatively charged is

• Ag⁺/Ag
• K⁺/K
• Any of the two
• None of them

Solution

Here K⁺/K is undergoing oxidation so it is anode and is negatively charged.

Q38. A cell is prepared by dipping a chromium rod in 1M Cr₂(SO₄)₃ solution and an iron rod in 1M FeSO₄ solution. The standard reduction potentials of Chromium and Iron electrodes are -0.75 V and -0.45 V respectively. What will be the standard EMF of the cell?

• - 0.30 V
• +0.30 V
• +0.47 V
• -0.47 V

Solution

The standard EMF of the cell, E_cell = E_cathode - E_anode So, Ecell = -0.45 - (-0.75) = 0.75-0.45 = +0.30 V.

Q39. Which of the following is a mild oxidizing agent?

• Ag₂O
• KMnO₄
• K₂Cr₂O₇
• Cl₂

Solution

Ag₂O is a mild oxidizing agent.

Q40. The reaction H₂S + H₂O₂ → SO₂ + 2H₂O manifests

• Oxidizing action of H₂O₂
• Reducing nature of H₂O₂
• Acidic nature of H₂O₂
• Alkaline nature of H₂O₂