Question

The standard potential of the electrode, Zn2+ (0.02M) Zn (s) is -0.76 V. Calculate its potential.​

Answer 1

Answer:

This question can be solved using Nernst equation and the answer is - 0.810 V

Explanation:

Using Nernst equation,

E = E° -0.0591/2 log 1/[Zn^2+]

E = -0.76 - 0.0591/2 log 1/0.02

E = -0.76-0.02955 log 50

E = -0.76 - 0.02955 x 1.6987 (log 50 = 1.6987)

E = -076 - 0.050

E = -0.810 V

where E is the potential of zinc electrode

Answer 2

The potential of the cell is -1.07 V.

Given:

The standard potential of the electrode reaction given is:

Zn2+ (0.02M) + 2e- → Zn (s) E° = -0.76 V

To find:

Potential of the cell having standard potential

Zn2+ (0.02M) + 2e- → Zn (s) E° = -0.76 V

Solution:

To calculate the potential of the cell, we need to use the Nernst equation:

E = E° - (RT/nF) ln(Q)

where:

E = potential of the cell

E° = standard potential of the cell

R = gas constant = 8.314 J/K.mol

T = temperature in Kelvin

n = number of electrons transferred in the cell reaction

F = Faraday's constant = 96485 C/mol

Q = reaction quotient

In this case, the reaction quotient Q can be calculated from the concentrations of the species involved in the cell reaction:

Q = [Zn]/([Zn2+]^(2))

Substituting the given values into the equation, we get:

Q = [Zn] / ([Zn2+]^2) = 1 / (0.02^2) = 2500

Now we can substitute all the values into the Nernst equation and calculate the potential of the cell:

E = E° - (RT/nF) ln(Q)

E = -0.76 V - [(8.314 J/K.mol) x (298 K) / (2 x 96485 C/mol)] x ln(2500)

E = -0.76 V - (0.0257 V) x ln(2500)

E = -0.76 V - 0.307 V

E = -1.07 V

Therefore, the potential of the cell is -1.07 V.

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