Answer to Question #166169 in Physical Chemistry for Tariq

Question

6. Calculate electromotive force of a galvanic element consisting of two silver

electrodes immersed in a solution of silver nitrate with active concentration of silver

ions 0.1 mol/1 and 0.01 mol/1, T=298 K.

A. E.M.F. = 0,049 V

B. E.M.F. = 0,059 V

C. E.M.F. = 0,069 V

Answer 1

The cell representation:

$Ag^{1+}|Ag^0||Ag^{1+}|Ag^0$

The cell reactions:

But first, you need to determine the cathode and anode. In the concentration element, the cathode is the electrode with the solution, the concentration of the solute of which is greater. And the electrode with the solution, the concentration of the dissolved substance of which is less, is the anode.

$A(-): Ag^0 = Ag^{1+} + 1e$

$C(+): Ag^{1+} + 1e = Ag^0$

The emf of the cell:

To find the EMF of a concentration element, you must use the following formula:

$E = \frac{RT}{zF} \times ln(\frac{c1}{c2})$

c1>c2

E = 0.08 V