The standard reduction potential for $C u^{2 +} / C u$ is $+ 0.34$ V. What will be the reduction potential at pH $= 14$ ? [Given: $K_{s p}$ of $C u (O H)_{2}$ is $1.0 \times 10^{- 19}]$ .

Question

Q. The standard reduction potential of

C u^{2 +} / C u

couple is

0.34

V at

25^{0}

C. Calculate the reduction potential at

p H = 14

for this couple.
Given

K_{s p}

of

C u (O H)_{2} = 1.0 \times 10^{- 18}

)

Answer 1

pH is given by,

p H = - l o g [H^{+}] = 14 \Rightarrow [H^{+}] = 10^{- 14} M

Hence,

[O H^{-}] = 1 M

since

[H^{+}] [O H^{-}] = 10^{- 14}; \Rightarrow [O H^{-}] = 1 M

C u (O H)_{2} \to C u^{2 +} + 2 O H^{-}

K_{s p} = [C u^{2 +}] [O H^{-}]^{2}

1 \times 10^{- 19} = [C u^{2 +}] \times 1 M; \Rightarrow [C u^{2 +}] = 1 \times 10^{- 19}

For the reaction,

C u^{2 +} + 2 e^{-} \to C u, n = 2

E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n}

l o g

\frac{1}{C u^{2 +}}

⟹ E_{c e l l} = 0.34 - \frac{0.0591}{2}

l o g

\frac{1}{1 \times 10^{- 19}} = - 0.22 V

Answer 2

pH is given by,

Answer 3

p H = - l o g [H^{+}] = 14 \Rightarrow [H^{+}] = 10^{- 14} M

Answer 4

Hence,

[O H^{-}] = 1 M

since

[H^{+}] [O H^{-}] = 10^{- 14}; \Rightarrow [O H^{-}] = 1 M

Answer 5

C u (O H)_{2} \to C u^{2 +} + 2 O H^{-}

Answer 6

K_{s p} = [C u^{2 +}] [O H^{-}]^{2}

1 \times 10^{- 19} = [C u^{2 +}] \times 1 M; \Rightarrow [C u^{2 +}] = 1 \times 10^{- 19}

Answer 7

For the reaction,

C u^{2 +} + 2 e^{-} \to C u, n = 2

Answer 8

E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n}

l o g

\frac{1}{C u^{2 +}}

Answer 9

⟹ E_{c e l l} = 0.34 - \frac{0.0591}{2}

l o g

\frac{1}{1 \times 10^{- 19}} = - 0.22 V

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].