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

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)