Question

Calculate the cell potential of the galvanic cell at 298К consists of two silver electrodes immersed into two solutions having с₁(Ag⁺)= 10 ^–2mol/L and с₂(Ag⁺)= 10 ^–1mol/L.

Answer 1

Cell potential E₀ = -0.059×log($\frac{c_1}{c_2}$ )

E₀ = -0.059× log($\frac{0.01}{0.1}$ )

E₀ = 0.059V