Question

You have constructed a quinhydrone concentration cell with buffers of pH = 4.2 and pH = 2.5 according to the following notation: C(s, graphite) | pH 4.7 buffer, quinhydrone || pH 2.7 buffer, quinhydrone | C(s, graphite) Using the Nernst equation as derived in the Chemical Foundations section for the quinhydrone concentration cell, calculate the cell potential for this electrochemical cell.

Answer 1

The the cell is

C (s,graphite)│pH 4.2 buffer, quinhydrone││ pH 2.5 buffer, quinhydrone │C (s, graphite)

The Nernst equations for the two half cells are given by

E = E⁰ + 0.0591 log₁₀[H⁺]

Now, we know that pH = -log₁₀[H⁺]; hence the equation can be re-written as

E = E⁰ – 0.0591 pH

The pH of the two half cells are given as 4.7 and 2.7

Therefore, E_cell = E_R – E_{L =} {E⁰ – 0.0591.(2.7)} – {E⁰ – 0.0591.(4.7)}

or, E_cell = 0.0591.(4.7 – 2.7) = 0.1182 V

Therefore, the cell potential is 0.1182 V