Question #265124

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.


1
Expert's answer
2021-11-13T01:33:47-0500

Cell potential E0 = -0.059×log(c1c2\frac{c_1}{c_2} )

E0 = -0.059× log(0.010.1\frac{0.01}{0.1} )

E0 = 0.059V


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