Question

The following chemical reaction is occurring in an electrochemical cell. Mg(s) + 2 Ag+ (0.0001 M) Mg2+ (0.10M) + 2 Ag(s) The electrode values are Mg2+ / Mg = – 2. 36 V Ag+ / Ag = 0.81 V For this cell calculate / write (a) (i) EO value for the electrode 2Ag+ / 2Ag (ii) Standard cell potential EO cell. (b) Cell potential (E)cell (c) (i) Symbolic representation of the above cell. (ii) Will the above cell reaction be spontaneous?​

Answer 1

this question answer is all details in pic

Answer 2

Answer: a) 1) $E^0_(2Ag^=/2Ag)$= 0.81 V

a 2) $E^0_{cell}=0.81-(-2.36)=3.17V$

b) $E_{cell}=2.96V$

c) (i)  $Mg/Mg^{2+}(0.10M)//Ag^+(0.0001M)/Ag$

ii) Yes it is spontaneous.

Explanation:-

a) (i) The electrode potential of the electrode is independent of the number of moles and hence it remains same for 1 mole.

Thus $E^0_(2Ag^=/2Ag)$= 0.81 V

a) (ii)  $Mg+2Ag^+\rightarrow Mg^{2+}+2Ag$

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

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

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

$E^0_{[Mg^{2+}/Mg]}= -2.36V$

$E^0_{[Ag^{+}/Ag]}= 0.81V$

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

$E^0=0.81-(-2.36)=3.17V$

b) Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Mg^{2+}]}{[Ag^+]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential = 3.17 V

$E_{cell}=3.17-\frac{0.0592}{2}\log \frac{[0.10]}{[0.0001]^2}$

$E_{cell}=2.96V$

c) 1) The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a slat bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

$Mg/Mg^{2+}(0.10M)//Ag^+(0.0001M)/Ag$

2) $E_{cell}$= +ve, reaction is spontaneous

$E_{cell}$= -ve, reaction is non spontaneous

$E_{cell}$= 0, reaction is in equilibrium

Thus as $E_{cell}=2.96V$ , the reaction is spontaneous.