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प्रश्न
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
उत्तर
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
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