Question

Calculate the standard cell potential for mg(s)+fe2+(aq)→mg2+(aq)+fe(s)

Answer 1

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Answer 2

Answer: 1.89 V

Explanation:- $Mg(s)+Fe^{2+}(aq)\rightarrow Mg^{2+}(aq)+Fe(s)$

Here Mg undergoes oxidation by loss of electrons, thus act as anode. Iron undergoes reduction by gain of electrons and thus act as cathode.

$EMF=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Mg^{2+}/Mg]}= -2.38V$

$E^0_{[Fe^{2+}/Fe]}= -0.49V$

$EMF=E^0_{[Fe^{2+}/Fe]}- E^0_{[Mg^{2+}/Mg]}$

$EMF=-0.49V- (-2.38V)=1.89V$