Question

Calculate the emf of the following cell at 298 K.

\[\ce{Cu/Cu^{2+}_{(0.025 M)}//Ag^+_{(0.005 M)}/Ag}\]

Given `"E"_("Cu"^(2+)//"Cu")^circ` = 0.34 V, `"E"_("Ag"^+//"Ag")^circ` = 0.80 V

1 Faraday = 96500 C mol ^-1

Answer 1

For the given reaction

\[\ce{Cu/Cu^{2+}_{(0.025 M)}//Ag^+_{(0.005 M)}/Ag}\]

Ag⁺/Ag act as cathode.

Cu/Cu²⁺ act as anode.

`"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`

= 0.80 - 0.34

= 0.46 V

The given reaction is 

\[\ce{Cu(s) + 2Ag^+_{ (aq)} -> Cu^{2+}_{ (aq)} + 2Ag}\]

From here n = 2

We can calculate the emf of cell by Nernst equation

`"E"_"cell" = "E"_"cell"^circ - 0.0591/"n" log  (["Cu"^(+ 2)])/(["Ag"^+]^2)`

`= 0.46 - 0.0591/2 log  ([0.025])/([0.005]^2)`

= 0.46 - 0.02995 log (1 × 10³)

= 0.46 - 0.02995 × 3

= 0.46 - 0.08985

= 0.37 V