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KCET · Chemistry · Electrochemistry

Consider the following 4 electrodes
\(\begin{aligned} & \mathrm{A}: \mathrm{Ag}^{+}(0.0001 \mathrm{M}) / \mathrm{Ag}(s) \\ & \mathrm{B}: \mathrm{Ag}^{+}(0.1 \mathrm{M}) / \mathrm{Ag}(s) \\ & \mathrm{C}: \mathrm{Ag}^{+}(0.01 \mathrm{M}) / \mathrm{Ag}(s) \\ & \mathrm{D}: \mathrm{Ag}^{+}(0.001 \mathrm{M}) / \mathrm{Ag}(s) ; E^{\circ}{ }_{\mathrm{Ag}^{+} / \mathrm{Ag}}=+0.80 \mathrm{~V}\end{aligned}\)
Then reduction potential in volts of the electrodes in the order.

  1. A \(\mathrm{B}>\mathrm{C}>\mathrm{D}>\mathrm{A}\)
  2. B \(C>D>A>B\)
  3. C \(A>D>C>B\)
  4. D \(A>B>C>D\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\mathrm{B}>\mathrm{C}>\mathrm{D}>\mathrm{A}\)

Step-by-step Solution

Detailed explanation


According to this equation, as concentration of metal ion increases, reduction potential of metal electrode also increases i.e \(E \propto\) concentration.
\(\therefore \quad\) B \(>\) C \(>\) D \(>\) A