1 Faraday electricity was passed through $$Cu^{2+}$$ (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through $$Ag^+$$ (0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is ______. (Given: $$E^0_{Cu^{2+}/Cu} = 0.34$$ V, $$E^0_{Ag^+/Ag} = 0.8$$ V, $$\frac{2.303RT}{F} = 0.06$$ V)

Initial data (volume of each solution = 1 L):

• $$[Cu^{2+}]_0 = 1.5\;M \Rightarrow n(Cu^{2+})_0 = 1.5\;mol$$
• $$[Ag^{+}]_0 = 0.2\;M \Rightarrow n(Ag^{+})_0 = 0.2\;mol$$

Step 1 : Effect of electrolysis on the $$Cu^{2+}/Cu$$ half-cell

Reaction at cathode: $$Cu^{2+}+2e^- \rightarrow Cu$$

1 F = 1 mol e⁻, therefore 1 F deposits $$\dfrac{1}{2}=0.5\;mol$$ of $$Cu^{2+}$$.

Remaining $$Cu^{2+}$$ moles: $$1.5-0.5=1.0\;mol$$
New concentration: $$[Cu^{2+}] = \dfrac{1.0}{1\;L}=1.0\;M$$

Step 2 : Effect of electrolysis on the $$Ag^{+}/Ag$$ half-cell

Reaction at cathode: $$Ag^{+}+e^- \rightarrow Ag$$

0.1 F = 0.1 mol e⁻, therefore 0.1 mol of $$Ag^{+}$$ is removed.

Remaining $$Ag^{+}$$ moles: $$0.2-0.1=0.1\;mol$$
New concentration: $$[Ag^{+}] = \dfrac{0.1}{1\;L}=0.1\;M$$

Step 3 : Construction of the galvanic cell

Standard reduction potentials: $$E^\circ_{Ag^{+}/Ag}=0.80\;V,\;E^\circ_{Cu^{2+}/Cu}=0.34\;V$$

Hence Ag⁺/Ag will act as the cathode and Cu/Cu²⁺ as the anode:

Cell: $$Cu(s)\,|\,Cu^{2+}(1.0\,M)\,||\,Ag^{+}(0.1\,M)\,|\,Ag(s)$$

Step 4 : Standard emf of the cell

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.80-0.34=0.46\;V$$

Step 5 : Emf under the given concentrations (Nernst equation)

Overall cell reaction: $$Cu(s)+2Ag^{+}\rightarrow Cu^{2+}+2Ag(s)$$

Number of electrons transferred, $$n=2$$.

Reaction quotient: $$Q=\dfrac{[Cu^{2+}]}{[Ag^{+}]^{2}}=\dfrac{1.0}{(0.1)^{2}}=100$$

Nernst equation (298 K):
$$E_{cell}=E^\circ_{cell}-\dfrac{0.0591}{n}\log Q$$
Using the given value $$\dfrac{2.303RT}{F}=0.06\;V$$:
$$E_{cell}=0.46-\dfrac{0.06}{2}\log 100$$

Since $$\log 100 = 2$$,
$$E_{cell}=0.46-(0.03)(2)=0.46-0.06=0.40\;V$$

Answer: The emf of the cell is $$0.40\;V$$, i.e. $$400\;mV$$.