Electrochemistry

Question

A zinc electrode is submerged in an acidic 0.80M Zn+ solution which are connected by salt bridge to a 1.30M Ag+ solution containing a silver electrode. Determine the initial voltage of the cell at 298k

Given that:

Zn/Zn2+ E°=0.76V

Ag+/Ag E°=0.80V

Answer 1

The cell can be written Zn(s) | Zn²⁺ (aq. 0.80 M) || Ag¹⁺ (aq 1.30M) | Ag (s)

The balanced reaction is Zn(s) + 2 Ag¹⁺ ====> Zn²⁺ + 2 Ag (s)

The cell potential can be found using the Nernst equation, which is

E_cell = E⁰_cell - RT/nF lnQ where R = 8.314 J/Kmole, F = 96487 coulombs/mole, Q = [Zn²⁺] / [Ag¹⁺]², n =2.

E⁰_cell = E⁰_Ag+/Ag - E⁰_Zn/Zn2+ = 0.80 V - 076 V = 0.04 V. Substitute these values into the Nernst equation.

E_cell = 0.04 - 8.314 x 298 /2 x 96487 x ln (0.80 / 1.3²) = 0.050 V

Always check math.