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Question: The standard reduction potential for \(Cu^{2 +}/Cu\) is \(+ 0.34V\) what will be the reduction poten...

The standard reduction potential for Cu2+/CuCu^{2 +}/Cu is +0.34V+ 0.34V what will be the reduction potential at pH = 14?

[Given KspK_{sp}g of Cu(OH)2is1.0×10−19Cu(OH)_{2}is1.0 \times 10^{- 19}]

A

2.2V2.2V

B

3.4V3.4V

C

−0.22V- 0.22V

D

−2.2V- 2.2V

Answer

−0.22V- 0.22V

Explanation

Solution

lpH=−log⁡[H+]=14pH = - \log\lbrack H^{+}\rbrack = 14

[H+]=10−14\lbrack H^{+}\rbrack = 10^{- 14}

Hence, [OH−]=1M\lbrack OH^{-}\rbrack = 1M [H+][OH−]=10−4\lbrack H^{+}\rbrack\lbrack OH^{-}\rbrack = 10^{- 4}

Cu(OH)2→Cu2++2OH−Cu(OH)_{2} \rightarrow Cu^{2 +} + 2OH^{-}

Ksp=[Cu2+][OH−]2K_{sp} = \lbrack Cu^{2 +}\rbrack\lbrack OH^{-}\rbrack^{2}

For the reaction Cu2++2e−→Cu,n=2Cu^{2 +} + 2e^{-} \rightarrow Cu,n = 2

Ecell=Eºcell−0.0591nlog⁡1[Cu2+]E_{cell} = Eº_{cell} - \frac{0.0591}{n}\log\frac{1}{\lbrack Cu^{2 +}\rbrack}

Ecell=0.34−0.05912log⁡11×10−19=−0.22VE_{cell} = 0.34 - \frac{0.0591}{2}\log\frac{1}{1 \times 10^{- 19}} = - 0.22V