Question

$Ag/AgCl_{(s)}, KCl(0.2M)||KBr(0.001M), AgBr_{(s)}/Ag$ calculate the EMF generated and assign correct polarity to each electrode for spontaneous process after taking into account the cell reaction at $25^{\circ}C$. $K_{sp}(AgCl) = 2.8 \times 10^{-10}$

$K_{sp}(AgBr) = 3.3 \times 10^{-13}$ [JEE Adv. 1992]

Answer 1

EMF = 0.0372 V, Left electrode (Ag/AgCl) is positive, Right electrode (Ag/AgBr) is negative

Answer 2

The given galvanic cell is $Ag/AgCl_{(s)}, KCl(0.2M)||KBr(0.001M), AgBr_{(s)}/Ag$. This is a concentration cell where the electrodes are Ag metal in contact with sparingly soluble silver halides and their corresponding halide ions.

Let's denote the left half-cell as L and the right half-cell as R.

1. Determine the standard reduction potentials for each half-cell: The general reduction reaction for a silver halide electrode is: $AgX(s) + e^- \rightleftharpoons Ag(s) + X^-(aq)$ The standard potential $E^\circ_{AgX/Ag}$ can be related to the standard potential of the $Ag^+/Ag$ electrode ($E^\circ_{Ag^+/Ag} = 0.80 V$) and the solubility product $K_{sp}(AgX)$ using the following relation: $E^\circ_{AgX/Ag} = E^\circ_{Ag^+/Ag} + \frac{0.0591}{1} \log K_{sp}(AgX)$ at $25^\circ C$.

For the Left Half-cell (Ag/AgCl): $K_{sp}(AgCl) = 2.8 \times 10^{-10}$ $E^\circ_{AgCl/Ag} = 0.80 + 0.0591 \log (2.8 \times 10^{-10})$ $\log (2.8 \times 10^{-10}) = \log 2.8 - 10 = 0.447 - 10 = -9.553$ $E^\circ_{AgCl/Ag} = 0.80 + 0.0591 \times (-9.553) = 0.80 - 0.5645 = 0.2355 V$

For the Right Half-cell (Ag/AgBr): $K_{sp}(AgBr) = 3.3 \times 10^{-13}$ $E^\circ_{AgBr/Ag} = 0.80 + 0.0591 \log (3.3 \times 10^{-13})$ $\log (3.3 \times 10^{-13}) = \log 3.3 - 13 = 0.5185 - 13 = -12.4815$ $E^\circ_{AgBr/Ag} = 0.80 + 0.0591 \times (-12.4815) = 0.80 - 0.7377 = 0.0623 V$

2. Calculate the actual electrode potentials using the Nernst equation: The Nernst equation for the reduction half-reaction $AgX(s) + e^- \rightleftharpoons Ag(s) + X^-(aq)$ is: $E_{AgX/Ag} = E^\circ_{AgX/Ag} - \frac{0.0591}{1} \log [X^-]$

For the Left Half-cell (Ag/AgCl): $[Cl^-] = 0.2 M$ $E_{Left} = 0.2355 - 0.0591 \log (0.2)$ $\log (0.2) = \log (2/10) = \log 2 - \log 10 = 0.301 - 1 = -0.699$ $E_{Left} = 0.2355 - 0.0591 \times (-0.699) = 0.2355 + 0.0413 = 0.2768 V$

For the Right Half-cell (Ag/AgBr): $[Br^-] = 0.001 M = 10^{-3} M$ $E_{Right} = 0.0623 - 0.0591 \log (10^{-3})$ $\log (10^{-3}) = -3$ $E_{Right} = 0.0623 - 0.0591 \times (-3) = 0.0623 + 0.1773 = 0.2396 V$

3. Determine the EMF generated and assign polarity: For a spontaneous process, the electrode with the higher reduction potential will act as the cathode (reduction), and the electrode with the lower reduction potential will act as the anode (oxidation). Comparing the calculated potentials: $E_{Left} = 0.2768 V$ $E_{Right} = 0.2396 V$ Since $E_{Left} > E_{Right}$, the left electrode (Ag/AgCl) will act as the cathode, and the right electrode (Ag/AgBr) will act as the anode.

The EMF of the cell is given by: $E_{cell} = E_{cathode} - E_{anode} = E_{Left} - E_{Right}$ $E_{cell} = 0.2768 V - 0.2396 V = 0.0372 V$

Polarity:

• Cathode (Left electrode, Ag/AgCl): Positive terminal
• Anode (Right electrode, Ag/AgBr): Negative terminal

4. Cell Reaction: At Anode (Right electrode, oxidation): $Ag(s) + Br^-(aq) \rightarrow AgBr(s) + e^-$ At Cathode (Left electrode, reduction): $AgCl(s) + e^- \rightarrow Ag(s) + Cl^-(aq)$ Overall spontaneous cell reaction: $AgCl(s) + Br^-(aq) \rightarrow AgBr(s) + Cl^-(aq)$