JEE Mains

1. Galvanic Cell Potential

• The cell potential for a galvanic cell is calculated using the formula:

$$E° _{cell} = E° _{cathode} - E° _{anode}$$

$$= (-0.44 V) - (-0.76 V)$$

$$= 0.32 V$$

Therefore, the cell potential for the galvanic cell is 0.32 V.

2. Electrolysis of CuSO4

• The mass of copper deposited at the cathode can be calculated using the formula:

$$m = \frac{I \times t \times M}{z \times F}$$

$$= \frac{2.5 A \times 30 \times 60 \times 63.55 g/mol}{2 \times 96500 C/mol}$$

$$= 0.24 g$$

Therefore, the mass of copper deposited at the cathode is 0.24 g.

3. Number of Moles of Electrons Transferred

• The number of moles of electrons transferred in a redox reaction can be calculated by multiplying the number of electrons transferred by the number of moles of the substance involved in the reaction.

• In this case, 3 moles of Sn2+ are oxidized, and each Sn2+ atom loses 2 electrons. Therefore, the total number of electrons transferred is 3 moles x 2 electrons/mole = 6 moles of electrons.

4. Titration of FeSO4 with KMnO4

• The balanced equation for the reaction between FeSO4 and KMnO4 is:

$$5FeSO4 + KMnO4 + H2SO4 → 5Fe2(SO4)3 + MnSO4 + K2SO4 + H2O$$

• From the balanced equation, we can see that 5 moles of FeSO4 react with 1 mole of KMnO4.

• Therefore, the volume of KMnO4 solution required to reach the equivalence point can be calculated as follows:

$$V_{KMnO4} = \frac{M_{FeSO4} \times V_{FeSO4}}{M_{KMnO4}}$$

$$= \frac{0.10 M \times V_{FeSO4}}{0.050 M}$$

$$= 2V_{FeSO4}$$

Therefore, the volume of KMnO4 solution required to reach the equivalence point is twice the volume of FeSO4 solution.

5. Concentration of H2O2 Solution

• The balanced equation for the reaction between H2O2 and KI is:

$$H2O2 + 2KI + 2H2SO4 → I2 + 2K2SO4 + 2H2O$$

• From the balanced equation, we can see that 1 mole of H2O2 reacts with 2 moles of KI. Therefore, the concentration of the H2O2 solution can be calculated as follows:

$$M_{H2O2} = \frac{M_{Na2S2O3} \times V_{Na2S2O3}}{2V_{H2O2}}$$

$$= \frac{0.10 M \times V_{Na2S2O3}}{2V_{H2O2}}$$

Therefore, we need twice the volume of H2O2 solution as the Na2S2O3 solution to react with it.

CBSE Board Exams

1. Oxidation and Reduction in terms of Electron Transfer

• Oxidation is a process that involves the loss of electrons.
• Reduction is a process that involves the gain of electrons.

2. Differences between Electrochemical and Electrolytic Cells

3. Types of Redox Reactions

• Combination reaction: Two or more substances combine to form a single product, accompanied by the evolution of heat and light. Example:

$$2Mg + O2 → 2MgO$$

• Decomposition reaction: A single compound breaks down into two or more simpler substances, usually with the input of heat or electricity. Example:

$$2H2O → 2H2 + O2$$

• Displacement reaction: A more reactive metal displaces a less reactive metal from its compound. Example:

$$Fe + CuSO4 → FeSO4 + Cu$$

• Redox reaction involving a change in oxidation state: In this type of reaction, the oxidation state of one or more elements changes. Example:

$$Zn + 2HCl → ZnCl2 + H2$$

4. Factors that Affect the EMF of a Galvanic Cell

• Standard reduction potentials of the half-cells
• Concentrations of the reactants and products
• Temperature

5. Calculation of Cell Potential

The cell potential for a galvanic cell can be calculated using the standard reduction potentials of the half-cells and the Nernst equation:

$$E° _{cell} = E° _{cathode} - E° _{anode}$$

$$E _{cell} = E° _{cell} - \frac{0.0592}{n} logQ$$

where:

• E° cell is the standard cell potential
• E cell is the cell potential under non-standard conditions
• E° anode is the standard reduction potential of the anode
• E° cathode is the standard reduction potential of the cathode
• Q is the reaction quotient

6. Function of a Salt Bridge

A salt bridge maintains electrical neutrality in an electrochemical cell by allowing the flow of ions between the half-cells.

7. Electrolysis

Electrolysis is a process that uses electrical energy to drive a chemical reaction that would not occur spontaneously. Examples of reactions that can be carried out by electrolysis include:

• Electrolysis of water: produces hydrogen and oxygen
• Electrolysis of brine (NaCl solution): produces chlorine gas and sodium hydroxide solution
• Electrolysis of molten aluminum oxide: produces aluminum metal and oxygen gas