The `EMF` of the following cellis `1.05V` at `25^(@)C:` `Pt,H_(2)(g)(1.0 atm)|NaOH(0.1m),NaCl(0.1M)|AgCl(s),Ag(s)` `a.` Write the cell reaction, `b.` Calculate `pK_(w)` of water.

The `EMF` of the following cellis `1.05V` at `25^(@)C:` `Pt,H_(2)(g)(1

1.	The `EMF` of the following cellis `1.05V` at `25^(@)C:` `Pt,H_(2)(g)(1.0 atm)\|NaOH(0.1m),NaCl(0.1M)\|AgCl(s),Ag(s)` `a.` Write the cell reaction, `b.` Calculate `pK_(w)` of water.
Answer» Correct Answer - `pH=8.63` Half cell reactions are `:` `[Zn(s) rarr Zn^(2+)(aq)+2e^(-)]` at anode `[H^(o+)+e^(-)rarr (1)/(2)H_(2)]` at cathode For zinc electrode `:` `E_(Znn^(2+))=E^(c-)._(Zn\|Zn^(2+))-(0.0591)/(n)log[(Zn^(2+))/(Zn(s))]` `=0.76-(0.0591)/(2)log .(0.1)/(1)` `=0.76-(0.0591)/(2)xx(-1)` `=+0.76+0.03=0.79V` Similarly, for hydrogen electrode `E_(H^(o+)\|H_(2))=0-(0.0591)/(2)log .(1)/([H^(o+)]^(2))` or `E_(H^(o+)\|H_(2))=0-(0.0591)/(2)xx(-log[H^(o+)])` `:. E_(H^(o+)\|H_(2))=-0.0591pH` Now, `,E_(cell)=E_(Zn\|Zn^(2+))-E_(H^(o+)\|H_(2))` `:.0.28=+0.79-0.0591pH` or `0.0591pH=+0.79-0.28` or ` 0.0591pH=0.51` `:.pH=(0.51)/(0.0591)=8.63`

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The `EMF` of the following cellis `1.05V` at `25^(@)C:` `Pt,H_(2)(g)(1.0 atm)|NaOH(0.1m),NaCl(0.1M)|AgCl(s),Ag(s)` `a.` Write the cell reaction, `b.` Calculate `pK_(w)` of water.

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