Question

Calculate the emf of the following cell at 298 K.
 $ \text{Fe}\left( \text{s} \right)\text{/F}{{\text{e}}^{\text{+2}}}\left( \text{0}\text{.001M} \right)\parallel {{\text{H}}^{\text{+}}}\left( \text{1M} \right)\text{/}{{\text{H}}_{\text{2}}}\left( \text{g} \right)\text{1bar, Pt} $ (Given, $ \text{E}_{\text{cell}}^{\text{0}}\text{ = + 0}\text{.44 V} $ ).

Answer 1

Hint: The reduction potential is defined as the tendency of the chemical species to acquire electrons from or lose electrons to an electrode and thereby get reduced or get oxidized. It can be calculated from the equilibrium constant.
Formula Used: $ {{\text{E}}_{{\text{cell}}}}{\text{ = }}{{\text{E}}^{\text{0}}}_{{\text{cell}}} - \dfrac{{{\text{0}}{\text{.0591}}}}{{\text{n}}}{\text{log }}{{\text{K}}_{\text{c}}} $
Where $ {{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{\left[ {{\text{F}}{{\text{e}}^{{\text{ + 2}}}}} \right]\left[ {{{\text{H}}_{\text{2}}}} \right]}}{{\left[ {{\text{Fe}}} \right]{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}^{\text{2}}}}} $

Complete step by step solution:
As per the position of the elements in the electrochemical series of the element that is lower in the series gets reduced while the one that appears higher in the series gets reduced. Here, iron should get oxidized and hydrogen should get reduced. As per that that, the cell reaction that follows is:
 $ \text{Fe (s) + }{{\text{H}}^{\text{+}}}\left( \text{aq} \right)\to \text{F}{{\text{e}}^{\text{+2}}}\text{ (aq) + }{{\text{H}}_{\text{2}}}\left( \text{g} \right) $
Putting the values in the Nernst equation we get,
 $ {{\text{E}}_{\text{cell}}}\text{= }{{\text{E}}^{\text{0}}}_{\text{cell}}-\dfrac{\text{0}\text{.0591}}{\text{n}}\text{log }{{\text{K}}_{\text{c}}} $
Where, n is the number of electrons transferred, $ {{\text{K}}_{\text{c}}} $ is the equilibrium constant of the reaction, which is equal to, $ {{\text{K}}_{\text{c}}}\text{=}\dfrac{\left[ \text{F}{{\text{e}}^{\text{+2}}} \right]\left[ {{\text{H}}_{\text{2}}} \right]}{\left[ \text{Fe} \right]{{\left[ {{\text{H}}^{\text{+}}} \right]}^{\text{2}}}} $
The concentration of the ferric solution is $ \text{0}\text{.001 M} $ while the amount of hydrogen gas that is liberated has a pressure that is equal to 1 bar.
Considering the concentration of metallic iron and hydrogen to be equal to 1, we get that, $ {{\text{K}}_{\text{c}}}\text{=}\dfrac{\left[ \text{F}{{\text{e}}^{\text{+2}}} \right]\times 1}{1\times {{\left[ {{\text{H}}^{\text{+}}} \right]}^{\text{2}}}}=\dfrac{0.001}{{{1}^{2}}} $ = $ 0.001 $
The cell potential is,
 $ \text{E}_{\text{cell}}^{\text{0}}\text{= E}_{\text{right}}^{\text{0}}\text{- E}_{\text{left}}^{\text{0}} $ = $ \text{ + 0}\text{.44 V} $
 Putting the value in the above equation, we get,
 $ {{\text{E}}_{\text{cell}}}\text{= 0}\text{.44 -}\dfrac{\text{0}\text{.0591}}{2}\text{log }\left( 0.001 \right) $
Or, $ {{\text{E}}_{\text{cell}}}\text{= 0}\text{.44 -}\dfrac{\text{0}\text{.0591}}{2}\text{log }\left( {{10}^{-3}} \right) $
Or, $ {{\text{E}}_{\text{cell}}}\text{= 0}\text{.44 + }\left( \text{3 }\!\!\times\!\!\text{ }\dfrac{\text{0}\text{.0591}}{\text{2}} \right)\text{= }\left( \text{0}\text{.44+0}\text{.089} \right)\text{ V} $
Or, $ {{\text{E}}_{\text{cell}}}\text{= 0}\text{.53 V} $
Hence the emf of the following cell at 298 K is equal to $ \text{0}\text{.53 V} $ .

Note:
The Nernst equation relates the reduction potential of an electrochemical reaction (half –cell or full cell) to the standard reduction potential, temperature and the activities (often approximated by the concentrations) of the chemical species that are undergoing reduction and oxidation. This reaction was named after Walther Nernst, a German chemical physicist who laid this equation.