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Redox Reactions and Electrochemistry: Chemistry | JEE Main

The logarithm of equilibrium constant for the reaction

\mathrm{Pd}^{2+} + 4\mathrm{Cl}^- \rightleftharpoons \mathrm{PdCl}_4^{2-}

is (Nearest integer)

\text{Given : } \frac{2.303\mathrm{RT}}{\mathrm{F}} = 0.06\text{ V}

\mathrm{Pd}_{(\mathrm{aq})}^{2+} + 2\mathrm{e}^- \rightleftharpoons \mathrm{Pd}(\mathrm{s}) \ \mathrm{E}^\ominus = 0.83\text{ V}

\mathrm{PdCl}_4^{2-}(\mathrm{aq}) + 2\mathrm{e}^- \rightleftharpoons \mathrm{Pd}(\mathrm{s}) + 4\mathrm{Cl}^-(\mathrm{aq}) \ \mathrm{E}^\ominus = 0.65\text{ V}

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Step-by-step solutionView

Correct answer

The logarithm of the equilibrium constant $\log_{10} K$ is 6.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Given half-cell reactions and potentials: (1) $\mathrm{Pd}^{2+} + 2\mathrm{e}^- \rightleftharpoons \mathrm{Pd}(\mathrm{s})$ , $E_1^\circ = 0.83\text{ V}$ (2) $\mathrm{PdCl}_4^{2-} + 2\mathrm{e}^- \rightleftharpoons \mathrm{Pd}(\mathrm{s}) + 4\mathrm{Cl}^-$ , $E_2^\circ = 0.65\text{ V}$ Also, $\frac{2.303RT}{F} = 0.06\text{ V}$ .

02find

Find the logarithm of the equilibrium constant, $\log_{10} K$ , for the target reaction: $\mathrm{Pd}^{2+} + 4\mathrm{Cl}^- \rightleftharpoons \mathrm{PdCl}_4^{2-}$

03visualise

The target reaction is obtained by subtracting reaction (2) from reaction (1): $\text{Target} = (1) - (2)$ $\mathrm{Pd}^{2+} + 4\mathrm{Cl}^- \rightleftharpoons \mathrm{PdCl}_4^{2-}$

04strategise

Calculate the cell potential for the target cell: $E^\circ = E_1^\circ - E_2^\circ$ Then use the thermodynamic equilibrium relationship: $E^\circ = \frac{2.303RT}{nF} \log_{10} K \implies \log_{10} K = \frac{n E^\circ}{0.06}$ where $n = 2$ is the number of electrons transferred in each half-reaction.

05execute

Compute $E^\circ = 0.83 - 0.65 = 0.18\text{ V}$ . Then: $\log_{10} K = \frac{2 \times 0.18}{0.06} = 6$

✓verify

Since $E^\circ > 0$ ( $0.18\text{ V}$ ), the formation of $\mathrm{PdCl}_4^{2-}$ from $\mathrm{Pd}^{2+}$ and $\mathrm{Cl}^-$ is thermodynamically spontaneous, so $K > 1$ and $\log K > 0$ . The integer value 6 is physically consistent.

Hints that build this answer step by step

How can the given half-reactions be combined to yield the target reaction $\mathrm{Pd}^{2+} + 4\mathrm{Cl}^- \rightleftharpoons \mathrm{PdCl}_4^{2-}$ ?
Reaction 1 minus Reaction 2
What is the standard cell potential $E^\ominus$ for this overall reaction?
$E^\ominus = 0.83\text{ V} - 0.65\text{ V} = +0.18\text{ V}$
Using $E^\ominus = \frac{2.303RT}{nF} \log_{10} K$ with $n = 2$ and $\frac{2.303RT}{F} = 0.06\text{ V}$ , what is the value of $\log_{10} K$ ?
6

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is n = 2 when no electrons appear in the overall net reaction?

The net complexation reaction is constructed from two half-reactions that each involve a 2-electron transfer ( $n = 2$ ). The electrons cancel in the net equation, but $n$ represents the electrons exchanged in the coupled redox processes.

Answer

The logarithm of the equilibrium constant $\log_{10} K$ is 6.

Why each option works or fails

Step-by-step solution

given: Given half-cell reactions and potentials: (1) $\mathrm{Pd}^{2+} + 2\mathrm{e}^- \rightleftharpoons \mathrm{Pd}(\mathrm{s})$ , $E_1^\circ = 0.83\text{ V}$ (2) $\mathrm{PdCl}_4^{2-} + 2\mathrm{e}^- \rightleftharpoons \mathrm{Pd}(\mathrm{s}) + 4\mathrm{Cl}^-$ , $E_2^\circ = 0.65\text{ V}$ Also, $\frac{2.303RT}{F} = 0.06\text{ V}$ .
find: Find the logarithm of the equilibrium constant, $\log_{10} K$ , for the target reaction: $\mathrm{Pd}^{2+} + 4\mathrm{Cl}^- \rightleftharpoons \mathrm{PdCl}_4^{2-}$
visualise: The target reaction is obtained by subtracting reaction (2) from reaction (1): $\text{Target} = (1) - (2)$ $\mathrm{Pd}^{2+} + 4\mathrm{Cl}^- \rightleftharpoons \mathrm{PdCl}_4^{2-}$
strategise: Calculate the cell potential for the target cell: $E^\circ = E_1^\circ - E_2^\circ$ Then use the thermodynamic equilibrium relationship: $E^\circ = \frac{2.303RT}{nF} \log_{10} K \implies \log_{10} K = \frac{n E^\circ}{0.06}$ where $n = 2$ is the number of electrons transferred in each half-reaction.
execute: Compute $E^\circ = 0.83 - 0.65 = 0.18\text{ V}$ . Then: $\log_{10} K = \frac{2 \times 0.18}{0.06} = 6$
verify: Since $E^\circ > 0$ ( $0.18\text{ V}$ ), the formation of $\mathrm{PdCl}_4^{2-}$ from $\mathrm{Pd}^{2+}$ and $\mathrm{Cl}^-$ is thermodynamically spontaneous, so $K > 1$ and $\log K > 0$ . The integer value 6 is physically consistent.