Question

At T(K), the electrode potential of A$^{2+}$(X M)$|$A(s) is –0.2285 V. Given $E^\circ_{A^{2+}/A} = -0.14$ V, the value of X in mol L$^{-1}$ is:

• A. 0.1
• B. 0.01
• C. 0.02
• D. 0.001

Hint:

Use the Nernst equation to relate concentration with electrode potential.

Answer 1

The Correct Option is D

Use Nernst equation: $E = E^\circ - \dfrac{0.0591}{n} \log \dfrac{1}{[A^{2+}]}$
Given: $E = -0.2285$ V, $E^\circ = -0.14$ V, $n = 2$
$-0.2285 = -0.14 - \dfrac{0.0591}{2} \log \dfrac{1}{X}$
$\Rightarrow \log \dfrac{1}{X} = \dfrac{-0.2285 + 0.14}{0.02955} \approx -2.99$
$\Rightarrow \log \dfrac{1}{X} = -3$ ⇒ $X = 10^{-3} = 0.001$ mol/L