Calculate the activity of KI, x, in the following electrochemical cell if the potential is +0.350 V. Ag|AgCl(s), NaCl(aq, a=0.1)||KI(aq, a=x), I2(s) | Pt

The activity KI for the given electrochemical cell is 0.0310

Explanation:

The given chemical cell follows:

Oxidation half reaction:

Reduction half reaction:

Net cell reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

To calculate the EMF of the cell, we use the Nernst equation, which is:

where,

= electrode potential of the cell = 0.350 V

= standard electrode potential of the cell = +0.32 V

n = number of electrons exchanged = 1

Putting values in above equation, we get:

Hence, the activity KI for the given electrochemical cell is 0.0310

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.