At 298 K, the standard reduction potential for \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) electrode is 0.34 V. Given: \(\mathrm{K}_{\mathrm{sp}} \mathrm{Cu}(\mathrm{OH})_{2}=1 \times 10^{-20}\) Take \(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\) The reduction potential at \(\mathrm{pH}=14\) for the above couple is \((-) \mathrm{x} \times 10^{-2} \mathrm{~V}\). The value of x is ____.

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Redox Reactions and Electrochemistry

At 298 K, the standard reduction potential for

Cu^{2 +} / Cu

electrode is 0.34 V.

Given:

K_{sp} Cu (OH)_{2} = 1 \times 1 0^{- 20}

Take

\frac{2.303 RT}{F} = 0.059 V

The reduction potential at

pH = 14

for the above couple is

(-) x \times 1 0^{- 2} V

. The value of x is ____.

Answer: 25

At pH = 14, pOH = 0, so [OH⁻] = 1 M. Using the solubility product:

K_{sp} = [Cu^{2 +}] [OH^{-}]^{2}

, we get

[Cu^{2 +}] = 1 0^{- 20}

M.

Applying the Nernst equation for the reduction

Cu^{2 +} + 2 e^{-} \to Cu

E = E^{\circ} - \frac{0.059}{2} lo g_{10} \frac{1}{[ Cu ^{2 +} ]}

E = 0.34 - \frac{0.059}{2} lo g_{10} (1 0^{20})

E = 0.34 - \frac{0.059 \times 20}{2} = 0.34 - 0.59 = - 0.25 V

Expressing as

(-) x \times 1 0^{- 2}

- 0.25 = - 25 \times 1 0^{- 2}

, so x = 25.

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer