Read the problem and list out the given quantity.
Use the formula:
E = Eo - $\frac{0.05916}{n}$ log10 $\frac{a_{Red}}{a_{Ox}}$
Where Eo = Standard Half cell reduction potential
z = No of moles transferred to the cell
ared =Chemical Activity for the reduction species
aox = Chemical Activity for the oxidation species
Substituting the values, we get the answer.
The Emf of the following cells are:
Mg(S)| Mg2+ (0.2 M) | | Ag+ (1 $\times$ 10-3)| Ag
Find the Reduction potential for this.
given : Eo Ag+/Ag = 1 V
Eo Mg2+/Mg = - 3V
Eo = Eo Ag+/Ag + Eo Mg2+/Mg = 1 + 3 = 4 V
Chemical Activity for the reduction species (ared) = 0.2
Chemical Activity for the oxidation species (aoxi) = 0.001.
Number of Moles of Electrons Transferred in the Cell (z) = 2.
The Cell reaction is Mg + 2Ag+ --> 2Ag + Mg2+.
E = Eo - $\frac{0.05916}{z}$ log10 $\frac{a_{Red}}{a_{Ox}}$
= 4 - $\frac{0.5916}{2}$ log10 $\frac{0.2}{0.001}$
= 4 - 0.02958 $\times$ 2.301.
= 4 - 0.068
= 3.932.
The Reduction Potential Eo = 3.932 V.
The Emf of the following cells are:
Zn| Zn2+ (1 M) | | Ag+ (1 M)| Ag
Find the Reduction potential for this if EoAg+/Ag= - 0.5 V and EoMg2+/Mg = + 1V.
given : Eo Ag+/Ag = - 0.5 V
Eo Mg2+/Mg = + 1V
Eo = Eo Ag+/Ag + Eo Mg2+/Mg = 0.5 + 1 = 1.5 V
Chemical Activity for the reduction species (ared) = 1
Chemical Activity for the oxidation species (aoxi) = 1.
Reduction Potential is
E = Eo - $\frac{0.05916}{z}$ log10 $\frac{a_{Red}}{a_{Ox}}$
= 1.5 - $\frac{0.5916}{2}$ log10 $\frac{1}{1}$
= 1.5 - 0.02958 $\times$ 0.
= 1.5 - 0.
= 1.5 V.
The Reduction Potential Eo = 1.50 V.