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Q. The oxidation potentials of following half-cell reactions are given $Zn \rightarrow Zn ^{2+}+2 e^{-} ; E^{\circ}=0.76\, V , Fe \rightarrow Fe ^{2+}+$ $2 e^{-} ; E^{\circ}=0.44\, V$ what will be the emf of cell, whose cell reaction is $Fe ^{2+}( aq .)+ Zn \rightarrow Zn ^{2+}( aq .)+ Fe$

Electrochemistry

Solution:

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$EMF =E_{\text {cathode }}-E_{\text {anode }}=0.44-(0.76)=+0.32\, V$