At 298 K, the standard reduction potential for Cu^2+/ Cu electrode is 0.034 V. Given: Ka, Cu(OH)2 = 1 10^-9. The reduction potential at pH = 14 for the above couple is (X) 10^-7. The value of X is:

Given: Standard reduction potential of Cu^2+/ Cu is 0.034 V and the solution works at pH = 14. Find: The value of X in the required form. From the dissolution equilibrium of copper hydroxide, Cu(OH)2 ( s) Cu^2+ ( aq) + 2 OH^- ( aq) we use K sp = [ Cu^2+] [ OH^-]^2 At pH = 14, we have pOH = 0, so [ OH^-] = 1 M Hence, [ Cu^2+] = K sp[ OH^-]^2 = 1 10^-201^2 = 1 10^-20 M For the reduction half-reaction Cu^2+ ( aq) + 2 e^- Cu ( s) the Nernst equation is E = E^ - 0.0592 10 ( 1[ Cu^2+]) Substituting the values, E = 0.034 - 0.0592 10 ( 11 10^-20) Since 10(10^20) = 20 we get E = 0.034 - 0.0592 20 E = 0.034 - 0.59 = -0.25 V Thus the reduction potential corresponds to 25 in the required form used in the solution. Therefore, the value of X is 25.

At 298 K, the standard reduction potential for Cu^2+/ Cu electrod… — Solution | JEE Chemistry

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At $298 \, \text{K}$ , the standard reduction potential for $\text{Cu}^{2+}/\text{Cu}$ electrode is $0.034 \, \text{V}$ . Given: $K_a$ , $\text{Cu(OH)}_2$ = $1 \times 10^{-9}$ . The reduction potential at $\text{pH} = 14$ for the above couple is $(X) \times 10^{-7}$ . The value of $X$ is:

Answer

Correct answer:25

Step-by-step solution

Standard Method

Given: Standard reduction potential of $\text{Cu}^{2+}/\text{Cu}$ is $0.034 \, \text{V}$ and the solution works at $\text{pH} = 14$ .

Find: The value of $X$ in the required form.

From the dissolution equilibrium of copper hydroxide,

\text{Cu(OH)}_2 (\text{s}) \rightleftharpoons \text{Cu}^{2+} (\text{aq}) + 2\text{OH}^- (\text{aq})

we use

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

At $\text{pH} = 14$ , we have $\text{pOH} = 0$ , so

[\text{OH}^-] = 1 \, \text{M}

Hence,

[\text{Cu}^{2+}] = \frac{K_{\text{sp}}}{[\text{OH}^-]^2} = \frac{1 \times 10^{-20}}{1^2} = 1 \times 10^{-20} \, \text{M}

For the reduction half-reaction

\text{Cu}^{2+} (\text{aq}) + 2\text{e}^- \rightarrow \text{Cu} (\text{s})

the Nernst equation is

E = E^\circ - \frac{0.059}{2} \log_{10} \left( \frac{1}{[\text{Cu}^{2+}]} \right)

Substituting the values,

E = 0.034 - \frac{0.059}{2} \log_{10} \left( \frac{1}{1 \times 10^{-20}} \right)

Since

\log_{10}(10^{20}) = 20

we get

E = 0.034 - \frac{0.059}{2} \times 20

E = 0.034 - 0.59 = -0.25 \, \text{V}

Thus the reduction potential corresponds to $25$ in the required form used in the solution. Therefore, the value of $X$ is $25$ .

Common mistakes

Using $[\text{OH}^-] = 10^{-14}$ at $\text{pH} = 14$ is incorrect because $\text{pH} = 14 \Rightarrow \text{pOH} = 0$ . Therefore, use $[\text{OH}^-] = 1 \, \text{M}$ instead.
Confusing the given equilibrium constant and substituting the wrong value in the solubility relation gives an incorrect $[\text{Cu}^{2+}]$ . First write $K_{\text{sp}} = [\text{Cu}^{2+}][\text{OH}^-]^2$ and then substitute carefully.
Applying the Nernst equation with the wrong sign is a common error. For the reduction reaction, the reaction quotient appears as $1/[\text{Cu}^{2+}]$ , so the logarithmic term must be subtracted as shown in the solution.

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