Question

A concentration cell is constructed using two NI electrodes with Ni2 concentration of 1. 0 m and 1. 0 x 10^-4 m in the two half-cells. Calculate the potential of the cell if the reaction is spontaneous

Answer 1

The potential of the concentration cell is 0.1182 V when the reaction is spontaneous.

What is Concentration Cell?

An electrolytic cell that contains two half cells with the same electrodes but with different concentrations is known as a concentration cell.

What is Nernst Equation?

Nernst Equation helps in determining the cell potential under the non-standard conditions that are at any known temperature, pressure, and concentration. It is given by

$E = E^o - \frac{2.303RT}{F} log Q$

At T = 298 K,

$E = E^o - \frac{0.0591}{n} logQ$

For Ni concentration cell, anode and cathode are the same and thus E° is zero

At anode,

$Ni (s)\rightarrow Ni^2^+ + 2e^-$

At Cathode,

$Ni^2^+ + 2e^-\rightarrow Ni (s)$

No of moles of electrons, n = 2

Then, the potential of the cell is given by

$E=- \frac{0.0591}{n} logQ$

$Q = \frac{[diluted Ni]}{[concentrated Ni]}$

diluted Ni = 1 x 10⁻⁴ M and concentrated Ni = 1 M

$E = \frac{-0.059}{2} log\frac{ 10^-^4}{1}\\ = \frac{-0.059}{2} \times -4 log10$

log(10) = 1

E = 0.1182 V

Hence, the potential of the cell is 0.1182 V.

Learn more about the Nernst equation:

brainly.com/question/15237843

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