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# NCERT

3.5    Write the Nernst equation and emf of the following cells at 298 K:

            Pt(s)|Br^{-}(0.010M)|Br_{2}(1)||H^{+}(0.030 M)| H_{2}(g)(I bar)Pt(s)

Answers (1)
D Devendra Khairwa

The nernst equation of the given reaction gives : 

                  E_{Cell} = E_{cell}^{\circ}\ - \frac{0.0591}{n}log \frac{1}{\left[Br^-]^2 [ H^+ \right ]^2}

or                           =(0-1.09)\ - \frac{0.0591}{2}\ log \frac{1}{(0.010)^2 (0.030)^2}

or                           =-1.09\ - 0.02955\times\ log(1.11\times10^7)

or                           =-1.09\ - 0.208 =\ -1.298\ V

So the required emf of the cell is -1.298 V.

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