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

           (ii) Fe(s)|Fe^{2+}(0.001M)||H^{+}(1M)|H_{2}(g)(1 bar)|Pt(s)

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The nernst equation for this gives :

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

This gives :                                        = 0 - (-0.44)- \frac{0.0591}{2}log \frac{0.001}{1^2}

or                                                      = 0.44 - 0.02955(-3) = 0.53\ V             

Thus the emf of the given galvanic cell is 0.53 V.

Posted by

Devendra Khairwa

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