At 298 ~K , the standard reduction potential for Cu^2+ / Cu electrode is 0.34 ~V . Given : K_sp Cu(OH)_2=1 \times 10^-20 Take \frac2.303 RTF=0.059 ~V The reduction potential at pH=14 for the above couple is (-) x \times 10^-2 ~V . The value of x is \underline\hspace2cm

Electrochemistry PYQ — JEE Main 2023 · 13 Apr · Shift 2 · Q19 with Solution

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Correct answer: 25

For the reduction,

\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}(s), \qquad E^\circ = 0.34\,\text{V}

At non-standard conditions, Nernst equation is:

E = E^\circ - \frac{0.059}{2} \log \frac{1}{[\mathrm{Cu}^{2+}]}

Since activity of solid Cu is 1.

This becomes

E = E^\circ + \frac{0.059}{2} \log [\mathrm{Cu}^{2+}]

Given:

K_{sp}(\mathrm{Cu(OH)}_2) = [\mathrm{Cu}^{2+}][\mathrm{OH}^-]^2 = 10^{-20}

At $\mathrm{pH}=14$ ,

\mathrm{pOH} = 0 \Rightarrow [\mathrm{OH}^-] = 1\,\text{M}

Hence,

[\mathrm{Cu}^{2+}] = \frac{10^{-20}}{(1)^2} = 10^{-20}

E = 0.34 + \frac{0.059}{2} \log(10^{-20})

Since

\log(10^{-20}) = -20

E = 0.34 + \frac{0.059}{2}(-20)

E = 0.34 - 0.0295 \times 20

E = 0.34 - 0.59 = -0.25\,\text{V}

Given potential is written as

(-)x \times 10^{-2}\,\text{V}

Now,

-0.25\,\text{V} = -25 \times 10^{-2}\,\text{V}

Therefore,

x = 25

Stored correct answer = 25, which matches the derived result.