The standard reduction potential for Cu2+/Cu is +0.34 V. What will be the reduction potential at pH = 14 ? [Given: Ksp of Cu(OH)2 is 1.0 × 10-19]

Question

The standard reduction potential for Cu2+/Cu is +0.34 V. What will be the reduction potential at pH = 14 ? [Given: Ksp of Cu(OH)2 is 1.0 × 10-19] (A) 2.2 V (B) 3.4 V (C) -0.22 V (D) -2.2 V

Tardigrade Admin · Accepted Answer

pH=-log[H+]=14 ⇒ [H+]=10-14 Hence, [OH-]=1 M [H+][OH-]=10-4 Cu(OH)2 → Cu2++2OH- Ksp=[Cu2+][OH-]2 For the reaction, Cu2++2e- → Cu, n=2 Ecell=E°cell=(0.0591/n) log (1/[Cu2+]) Ecell=0.34-(0.0591/2) log (1/1×10-19) =-0.22 V

Q. The standard reduction potential for $C u^{2 +} / C u$ is $+ 0.34 V$ . What will be the reduction potential at $p H = 14$ ? [Given: $K_{s p}$ of $C u (O H)_{2}$ is $1.0 \times 1 0^{- 19}$ ]

$2.2 V$

$3.4 V$

$- 0.22 V$

$- 2.2 V$

Q. The standard reduction potential for Cu2+/Cu is +0.34V. What will be the reduction potential at pH=14 ? [Given: Ksp​ of Cu(OH)2​ is 1.0×10−19]

2.2V

3.4V

−0.22V

−2.2V

pH=−log[H+]=14 ⇒[H+]=10−14 Hence, [OH−]=1M[H+][OH−]=10−4 Cu(OH)2​→Cu2++2OH− Ksp​=[Cu2+][OH−]2 For the reaction, Cu2++2e−→Cu, n=2 Ecell​=Ecell∘​=n0.0591​log[Cu2+]1​ Ecell​=0.34−20.0591​log1×10−191​ =−0.22V

Q. The standard reduction potential for $C u^{2 +} / C u$ is $+ 0.34 V$ . What will be the reduction potential at $p H = 14$ ? [Given: $K_{s p}$ of $C u (O H)_{2}$ is $1.0 \times 1 0^{- 19}$ ]

$2.2 V$

$3.4 V$

$- 0.22 V$

$- 2.2 V$