**1. A] Define Single Electrode Potential. Derive Nernst equation for single electrode potential.**

**Answer:**

**SINGLE ELECTRODE POTENTIAL:** “Single electrode potential is defined as the potential generated when the metal is dipped in the solution consisting of its own ions, at the interphase between solution and metal”

**NERNST EQUATION: **

Nernst equation is a thermodynamic equation which relates the cell potential with concentrations Mn+ using standard free energy equation.

The decrease in free energy change (-∆G) is given by the maximum amount of work done by an electrochemical cell.−∆𝐺=𝑊_{𝑚𝑎𝑥}−−−→1

The maximum work done by the electrochemical cell depends on, Number of coulombs that flow and the energy available per coulomb.

𝑊_{𝑚𝑎𝑥}=𝑁𝑜 𝑜𝑓 𝑐𝑜𝑙𝑜𝑚𝑏𝑠 𝑋 𝐸𝑛𝑒𝑟𝑔𝑦 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑝𝑒𝑟 𝐶𝑜𝑢𝑙𝑜𝑚𝑏

The number of coulombs that flow is equal to the number of moles of electrons (n) and the faraday(F).

∴𝑁𝑜 𝑜𝑓 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 = 𝑛𝐹

Energy available per coulomb is the emf of the cell E. The maximum work done for an electrochemical cell is given by

𝑊_{𝑚𝑎𝑥}=𝑛𝐹𝑋𝐸——-2

Substituting equation 2, in 1 we have,

∆𝑮=−𝒏𝑭𝑬−−−→3

When the concentrations of all species is unity at 250C the standard free energy change ∆𝐺_{0} is given as

∆𝑮_{𝟎}=−𝒏𝑭𝑬_{𝟎}−−−→4

Where E_{0} is the standard electrode potential

“Standard electrode potential is the potential when a metal is dipped in 1M solution of its ions or when an inert electrode is in contact with a gas at temperature at 298K”

Consider a red-ox reaction involved in an electrochemical cell,

The equilibrium constant K_{c} is given by change in free energy by the equation,

∆𝐺=∆𝐺_{0}+𝑅𝑇𝑙𝑛𝐾_{𝑐} 𝑤ℎ𝑒𝑟𝑒 𝐾_{𝑐}=[𝑀][𝑀𝑛+]

Therefore, the above equation becomes,

Substituting equations 3 and 4 in equation 5 we have,

−𝑛𝐹𝐸 = −𝑛𝐹𝐸_{0} + 𝑅𝑇𝑙𝑛[𝑀] − 𝑅𝑇𝑙𝑛[𝑀𝑛+]

Dividing throughout by –nF, and under standard conditions M=1. Hence the above equation becomes,

Nernst equation at 298K and converting natural log to the base 10 is,