For pH = 14, [H⁺]= 10^-14
\([OH^-]={K_w\over [H^+]}={10^{-14}\over 10^{-14}}=1M\)
Cu(OH)₂ ⇌ Cu²⁺+ 2OH^-
K_sp= [Cu²⁺] [OH^-]²
1.0 x 10^-19 = [Cu²⁺](1)²
[Cu²⁺] = 1.0 X 10^-19
For the cell reaction,
Cu²⁺+ 2e^- ⟶ Cu
\(E=E^0{0.059\over 2}log{1\over [Cu^{2+}]}\)
\(=0.34-{0.059\over 2}log{1\over 1.0\times 10^{-19}}\)
\(=0.34-{0.059\over 2}\times19=-0.22V\)

\(For \ water \ at \ 298K,[{ H }^{ + }]={ 10 }^{ -7 }M\)
\(Reduction \ reaction: \ { H }^{ + }+{ e }^{ - }\longrightarrow \frac { 1 }{ 2 } { H }_{ 2 }\)
\( \therefore E=-\frac { RT }{ F } In\frac { { p }_{ { H }_{ 2 } }^{ 1/2 } }{ [{ H }^{ + }] } =-0.0591log\frac { { p }_{ { H }_{ 2 } }^{ 1/2 } }{ [{ H }^{ + }] } \)
\(=-.0.0591log\frac { 1 }{ { 10 }^{ -7 } } =-0.4137=-0.414V\)

1 mol = \(107.9g\)
1 mol of Ag is deposit by \(96500C\)
If charge of electron is \(1.60\times { 10 }^{ -19 }C\) 
No. of electrons in 1 mol of silver = \(\frac { 96500C }{ 1.60\times { 10 }^{ -19 }C } \) 
= \(6.03\times { 10 }^{ 23 }{ e }^{ - }\)
Avagadro number = \(6.03\times { 10 }^{ 23 }\)

(i) \({ Br }_{ 2 }+2{ I }^{ - }\longrightarrow 2{ Br }^{ - }+{ I }_{ 2 }\)
\({ E }^{ \theta }={ E }^{ \theta }\left( \frac { 1 }{ 2 } { Br }_{ 2 }|{ Br }^{ - } \right) -E°\left( \frac { 1 }{ 2 } { I }_{ 2 }|{ I }^{ - } \right)\)
\(=1.09-0.54=0.55V\) 
(ii) \(Pt|{ I }_{ 2 },{ I }^{ - }(aq)||{ Br }^{ - }(aq){ Br }_{ 2 }|Pt\) 
(iii) The value of \({ E }^{ \theta }\) will be reduced.

\(\Delta G=\_ ve,K=+ve,{ E° }_{ cell }=+ve\)