Question

Calculate E.M.F. of following cell at 298 K Zn_(s) |ZnSO₄ (0.01 M)| |CuSO₄ (1.0 M)| Cu_(s) if \[\ce{E^0_{cell}}\] = 2.0 V.

Options

• 2.0 V

• 2.0592 V

• 2.0296 V

• 1.0508 V

Answer 1

2.0592 V

Explanation:

General form of Nernst equation is

`"E"_("M"^"n+"//"M") = "E"_("M"^("n"+//"M"))^0 - "RT"/"nF" ln  ["M"]/(["M"^("n"+)])`

Oxidation - \[\ce{Zn(s) -> Zn^{2+} (aq) [0.01 M] + 2e-}\]

Reduction - \[\ce{Cu^{2+}(aq) [1.0] M -> Cu(s)}\]

Using Nernst equation,

`"E"_"cell" = "E"_"cell"^0 - 0.059/2 log  [1/100]`

`= 2 - 0.59/2 xx - 2 = 2 + 0.059`

= 2.059 V