31 public questions tagged with this topic.
2IO₃⁻ + 12H⁺ + 10e⁻ → I₂ + 6H₂O . For 1 mol IO₃⁻ , 5 electrons.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
E°cell = (0.059/n) log Kc , 1.36 = (0.059/n) × 46 , n = (2.714/1.36) ≈ 2 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
E°cell = (0.059/n) log Kc , 1.36 = (0.059/n) × 46 , n = (2.714/1.36) ≈ 2 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
NO₃⁻ + 10H⁺ + 8e⁻ → NH₄⁺ + 3H₂O . 8 electrons per mole.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Fe³⁺ + 3e⁻ → Fe . 3 electrons are required per mole.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Fe³⁺ + 3e⁻ → Fe . 3 electrons are required per mole.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
ClO₄⁻ + 8H⁺ + 8e⁻ → Cl⁻ + 4H₂O . 8 electrons per mole.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
In SO₃ + S → 2SO₂, SO₃ (S: +6 to +4) loses oxygen and is reduced, oxidizing S (0 to +4), no metal involved.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Redox Titrations and Equivalent Weight and Normality
Bi (+3 in Bi₂O₃) becomes 0 in Bi. Change = 0 - (+3) = -3; each Bi gains 3 electrons.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Redox Titrations and Equivalent Weight and Normality
I (0 in I₂) becomes +5 in HIO₃. Per I atom: 0 to +5 = 5 electrons.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Redox Titrations and Equivalent Weight and Normality
Oxidation: 10Cl− → 5Cl₂ + 10e−. Reduction: 2MnO₄− + 10e− → 2Mn₂+. Total electrons = 10.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Redox Titrations and Equivalent Weight and Normality
Oxidation: 6I− → 3I₂ + 6e−. Reduction: 2MnO₄− + 6e− → 2MnO₂. Total electrons = 6.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Redox Titrations and Equivalent Weight and Normality
