Question

What is the value of ΔGo in kJ at 25 oC for the reaction between the pair: Mn(s) and Ag+(aq) to give Ag(s) and Mn2+(aq) Use the reduction potential values for Ag+(aq) of +0.80 V and for Mn2+(aq) of -1.18 V Give your answer using E-notation with ONE decimal place

Answer 1

Answer:  $-3.8\times 10^{5}J$

Explanation:

$Mn+2Ag^{+}\rightarrow Mn^{2+}+2Ag$

Here Mn 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.

$E^0_{[Mn^{2+}/Mn]}= -1.18V$

$E^0_{[Ag^{2+}/Ag]}=+0.80V$

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

$E^0=+0.80- (-1.18V)=1.98V$

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G^0=-nFE^0$

$\Delta G^0$ = gibbs free energy

n= no of electrons gained or lost  = 2

F= faraday's constant

$E^0$ = standard emf  = 1.98V

$\Delta G^0=-2\times 96500\times (1.98)=-3.8\times 10^{5}J$

Thus the value of $\Delta G^0$ is $-3.8\times 10^{5}J$