Name the type of cell which exert EMF, though both electrodes are same

1.	Name the type of cell which exert EMF, though both electrodes are same but differ in electrolyte concentration. Two Ag electrodes immersed in \( Ag _{2} SO _{4} \) solution of \( 0.02 M \) and \( 0.042 \) \( M \) at \( 298 K \). Write the cell representation, cell reaction and calculate EMF of the cell.
Answer» This type of cell is known as electrolyte concentration cell. For this type of cell\(E^0_{cell} =0.\) At anode \(2Ag(s) \longrightarrow \underset{c_1 = 0.02 M} {2Ag^+}(aq) + 2e^- \; \text{oxidation}\) At cathode \(\underset{c_2 = 0.042 m}{Ag^\oplus_2SO_4}(aq) + 2e^-\longrightarrow 2Ag(s) + SO_4 ^{-2}(aq) \;\text{reduction}\) Cell representation- \(Ag(s)/Ag_2SO_4(0.02M) // Ag_2SO_4(0.042M)/Ag(s)\) Overall cell reaction- \(Ag^\oplus_2 SO_4 (C_2) \longrightarrow Ag_2SO_4 (C_1)\) or\(Ag^+(c_2) \longrightarrow Ag^+ (c_1)\) \(\therefore E_{cell} = E^0_{cell} -\frac{RT}{nf}ln \frac{c_1}{c_2}\) ⇒\(E_{cell} = 0 - \frac{0.0591}1log\frac{0.02}{0.042}\) \(\therefore E_{cell} = -0.0591 \times log(0.4762)\) \(= -0.0591 \times (-0.322)\) \(= 0.019\) \(E_{cell} = 0.02V\) \(E_{cell} = 20\, mV\)

Name the type of cell which exert EMF, though both electrodes are same but differ in electrolyte concentration. Two Ag electrodes immersed in \( Ag _{2} SO _{4} \) solution of \( 0.02 M \) and \( 0.042 \) \( M \) at \( 298 K \). Write the cell representation, cell reaction and calculate EMF of the cell.

Answer»

This type of cell is known as electrolyte concentration cell. For this type of cell\(E^0_{cell} =0.\)

At anode

\(2Ag(s) \longrightarrow \underset{c_1 = 0.02 M} {2Ag^+}(aq) + 2e^- \; \text{oxidation}\)

At cathode

\(\underset{c_2 = 0.042 m}{Ag^\oplus_2SO_4}(aq) + 2e^-\longrightarrow 2Ag(s) + SO_4 ^{-2}(aq) \;\text{reduction}\)

Cell representation-

\(Ag(s)/Ag_2SO_4(0.02M) // Ag_2SO_4(0.042M)/Ag(s)\)

Overall cell reaction-

\(Ag^\oplus_2 SO_4 (C_2) \longrightarrow Ag_2SO_4 (C_1)\)

or\(Ag^+(c_2) \longrightarrow Ag^+ (c_1)\)

\(\therefore E_{cell} = E^0_{cell} -\frac{RT}{nf}ln \frac{c_1}{c_2}\)

⇒\(E_{cell} = 0 - \frac{0.0591}1log\frac{0.02}{0.042}\)

\(\therefore E_{cell} = -0.0591 \times log(0.4762)\)

\(= -0.0591 \times (-0.322)\)

\(= 0.019\)

\(E_{cell} = 0.02V\)

\(E_{cell} = 20\, mV\)