the emf of a cell M/Mⁿ⁺ (0.02 M) // H⁺ (1 M )/H₂(g) (1atm), Pt at 25^o C is 0.81 V. then the valency of the metal is if the std. oxidation potential of the metal is 0.76

To determine the valency of the metal in the given electrochemical cell, we can use the Nernst equation, which relates the electromotive force (emf) of a cell to the standard electrode potential and the concentrations of the reactants and products involved in the half-reactions. Let's break this down step by step.

Understanding the Components of the Cell

The cell notation provided is M/Mⁿ⁺ (0.02 M) // H⁺ (1 M)/H₂ (g) (1 atm), Pt. Here, M represents the metal, and Mⁿ⁺ indicates its ion form. The concentration of the metal ion is 0.02 M, while the concentration of hydrogen ions is 1 M, and the hydrogen gas is at 1 atm pressure.

Standard Electrode Potential

The standard oxidation potential of the metal is given as 0.76 V. This means that when the metal is oxidized to its ion form, it has a tendency to lose electrons, which is reflected in the positive potential value.

Applying the Nernst Equation

The Nernst equation is expressed as:

E = E_° - (RT/nF) ln(Q)

Where:

• E = cell potential (0.81 V)
• E_° = standard cell potential (0.76 V for the metal)
• R = universal gas constant (8.314 J/(mol·K))
• T = temperature in Kelvin (25°C = 298 K)
• n = number of moles of electrons transferred (valency of the metal)
• F = Faraday's constant (96485 C/mol)
• Q = reaction quotient

Calculating the Reaction Quotient (Q)

For the half-reaction of the metal, we can express Q as:

Q = [Mⁿ⁺]/[H⁺]

Substituting the known concentrations:

Q = (0.02)/(1) = 0.02

Substituting Values into the Nernst Equation

Now, we can rearrange the Nernst equation to solve for n:

0.81 = 0.76 - (8.314 × 298)/(n × 96485) ln(0.02)

First, calculate the term involving R, T, and F:

(8.314 × 298)/(96485) ≈ 0.008314 V

Now, substituting this back into the equation:

0.81 = 0.76 - (0.008314/n) ln(0.02)

Rearranging gives:

0.81 - 0.76 = - (0.008314/n) ln(0.02)

0.05 = - (0.008314/n) ln(0.02)

Calculating ln(0.02)

Using a calculator, we find:

ln(0.02) ≈ -3.912

Substituting this value back into the equation:

0.05 = (0.008314 × 3.912)/n

Now, solving for n:

n = (0.008314 × 3.912)/0.05

n ≈ 0.651

Determining the Valency

Since n represents the number of electrons transferred in the oxidation process, we round this value to the nearest whole number. Thus, the valency of the metal is approximately 1.

In summary, based on the provided data and calculations, the valency of the metal in the electrochemical cell is 1. This indicates that the metal can lose one electron during the oxidation process to form its ion.