Question

$Ag(s)|Ag^+(aq)(1M)||Ag^+(aq)(2M)|Ag(s)$

(1L) (1L)

0.4 F of electricity in anode and 0.8 F in the cathode is first passed by making them independent electrolytic cell at 298 K. EMF of cell after electrolysis will be

Answer 1

-0.0040 V

Answer 2

The problem describes a concentration cell and then asks for its EMF after some electrolysis is performed independently on each compartment.

1. Initial State of the Cell: The cell is given as: $Ag(s)|Ag^+(aq)(1M)||Ag^+(aq)(2M)|Ag(s)$ This implies:

• Anode (Left Compartment): $Ag(s) \rightarrow Ag^+(aq) + e^-$ with initial $[Ag^+] = 1 \text{ M}$. Volume = 1 L, so initial moles of $Ag^+$ = 1 mole.
• Cathode (Right Compartment): $Ag^+(aq) + e^- \rightarrow Ag(s)$ with initial $[Ag^+] = 2 \text{ M}$. Volume = 1 L, so initial moles of $Ag^+$ = 2 moles.

2. Effect of Electrolysis:

• Anode Compartment (Independent Electrolytic Cell): 0.4 F of electricity is passed. In an electrolytic cell, the anode is where oxidation occurs. Reaction at anode: $Ag(s) \rightarrow Ag^+(aq) + e^-$ 1 Faraday (F) of electricity corresponds to 1 mole of electrons. So, 0.4 F means 0.4 moles of electrons are passed. According to the stoichiometry, 1 mole of electrons produces 1 mole of $Ag^+$. Therefore, 0.4 moles of electrons will produce 0.4 moles of $Ag^+$. Initial moles of $Ag^+$ in the anode compartment = $1 \text{ M} \times 1 \text{ L} = 1 \text{ mole}$. Final moles of $Ag^+$ in the anode compartment = $1 \text{ mole} + 0.4 \text{ moles} = 1.4 \text{ moles}$. Since the volume is 1 L, the final concentration of $Ag^+$ at the anode ($[Ag^+]_L$) = $1.4 \text{ M}$.

• Cathode Compartment (Independent Electrolytic Cell): 0.8 F of electricity is passed. In an electrolytic cell, the cathode is where reduction occurs. Reaction at cathode: $Ag^+(aq) + e^- \rightarrow Ag(s)$ 0.8 F means 0.8 moles of electrons are passed. According to the stoichiometry, 1 mole of electrons consumes 1 mole of $Ag^+$. Therefore, 0.8 moles of electrons will consume 0.8 moles of $Ag^+$. Initial moles of $Ag^+$ in the cathode compartment = $2 \text{ M} \times 1 \text{ L} = 2 \text{ moles}$. Final moles of $Ag^+$ in the cathode compartment = $2 \text{ moles} - 0.8 \text{ moles} = 1.2 \text{ moles}$. Since the volume is 1 L, the final concentration of $Ag^+$ at the cathode ($[Ag^+]_R$) = $1.2 \text{ M}$.

3. EMF of the Cell After Electrolysis: The cell after electrolysis is: $Ag(s)|Ag^+(aq)(1.4M)||Ag^+(aq)(1.2M)|Ag(s)$ This is a concentration cell, so the standard cell potential ($E^\circ_{cell}$) is 0 V. The Nernst equation for a concentration cell at 298 K is: $E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q$ Where:

• $n$ is the number of electrons transferred in the cell reaction ($n=1$ for $Ag^+/Ag$).
• $Q$ is the reaction quotient. For the cell reaction $Ag^+(aq)_{cathode} \rightarrow Ag^+(aq)_{anode}$, $Q = \frac{[Ag^+]_{anode}}{[Ag^+]_{cathode}}$.

Substituting the values: $[Ag^+]_{anode} = 1.4 \text{ M}$ $[Ag^+]_{cathode} = 1.2 \text{ M}$

$E_{cell} = 0 - \frac{0.0592}{1} \log \left(\frac{1.4}{1.2}\right)$ $E_{cell} = -0.0592 \log \left(\frac{7}{6}\right)$ $E_{cell} = -0.0592 (\log 7 - \log 6)$ Using $\log 7 \approx 0.8451$ and $\log 6 \approx 0.7782$: $E_{cell} = -0.0592 (0.8451 - 0.7782)$ $E_{cell} = -0.0592 (0.0669)$ $E_{cell} \approx -0.003959 \text{ V}$

Rounding to appropriate significant figures, $E_{cell} \approx -0.0040 \text{ V}$. The negative sign indicates that the cell reaction, as written (oxidation at the left electrode, reduction at the right electrode), is non-spontaneous. The spontaneous reaction would be in the reverse direction.

The final answer is $\boxed{-0.0040 \text{ V}}$

Explanation of the solution:

1. Calculate final concentrations:
   • Anode compartment: Initial moles $Ag^+$ = 1 M $\times$ 1 L = 1 mol. Electrolysis adds 0.4 mol $Ag^+$ ($0.4 \text{ F} \times 1 \text{ mol } Ag^+/\text{F}$). Final $[Ag^+]$ = (1 + 0.4) mol / 1 L = 1.4 M.
   • Cathode compartment: Initial moles $Ag^+$ = 2 M $\times$ 1 L = 2 mol. Electrolysis consumes 0.8 mol $Ag^+$ ($0.8 \text{ F} \times 1 \text{ mol } Ag^+/\text{F}$). Final $[Ag^+]$ = (2 - 0.8) mol / 1 L = 1.2 M.
2. Apply Nernst Equation: For a concentration cell $E^\circ_{cell} = 0$. The cell is $Ag(s)|Ag^+(aq)(1.4M)||Ag^+(aq)(1.2M)|Ag(s)$. $E_{cell} = -\frac{0.0592}{n} \log \left(\frac{[Ag^+]_{anode}}{[Ag^+]_{cathode}}\right)$ $E_{cell} = -\frac{0.0592}{1} \log \left(\frac{1.4}{1.2}\right)$ $E_{cell} = -0.0592 \log \left(\frac{7}{6}\right) \approx -0.0592 \times 0.0669 \approx -0.0040 \text{ V}$.