At 298 K, the standard reduction potential for the Cu 2 + / Cu electrode is 0.34 V. Given: K s p Cu(OH) 2 = 1 × 10 - 20 Take 2 . 303 R T F = 0 . 059 V The reduction potential at pH = 14 for the above couple is ( - ) x × 10 - 2 V. The value of x

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Solution

Q.
At 298 K, the standard reduction potential for the ${Cu}^{2 +} / Cu$ electrode is 0.34 V.

$Given: K_{s p} {Cu(OH)}_{2} = 1 \times 10^{- 20}$

$Take \frac{2.303 R T}{F} = 0.059 V$

The reduction potential at pH = 14 for the above couple is $(-) x \times 10^{- 2}$ V. The value of $x$ is __________ . [2023]

Ans.
(25)

$pH = 14 \Rightarrow pOH = 0$

$[{OH}^{-}] = 1$

$K_{s p} {Cu(OH)}_{2} = [{Cu}^{2 +}] {[{OH}^{-}]}^{2} = 1 \times 10^{- 20}$

$[{Cu}^{2 +}] = 10^{- 20}$

$E_{{Cu}^{2 +} | Cu} = E_{{Cu}^{2 +} | Cu}^{0} - \frac{0.059}{2} \log \frac{1}{[{Cu}^{2 +}]}$

$= 0.34 - \frac{0.059}{2} \log 10^{20}$

$= - 0.25$

$= - 25 \times 10^{- 2} V$

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