• # question_answer For the cell $Pt|{{H}_{2}}(g)|$ solution $X||KCI$ (saturated) $|H{{g}_{2}}C{{l}_{2}}|Hg|\,Pt$ the observed EMF at $25{}^\circ C$ was $600\text{ }mV.$ When solution X was replaced by a standard phosphate buffer with $pH=7.00,$ the EMF was $777\text{ }mV$. Find the pH of solution X. A) 3 B) 4            C) 5 D) 6

 [b] ${{H}_{2}}\xrightarrow{{}}2{{H}^{+}}+2{{e}^{-}}$ $\frac{H{{g}_{2}}C{{l}_{2}}+2{{e}^{-}}\xrightarrow{{}}2Hg+2C{{l}^{-}}}{{{H}_{2}}+H{{g}_{2}}C{{l}_{2}}\xrightarrow{{}}2{{H}^{+}}+2C{{l}^{-}}+2Hg}$ ${{E}_{initial}}=E{}^\circ +\frac{0.059}{2}\log \frac{1}{{{[{{H}^{+}}]}^{2}}\,{{[C{{l}^{-}}]}^{2}}}=0.6$ ${{E}_{\text{final}}}=E{}^\circ +\frac{0.059}{2}\log \frac{1}{{{({{10}^{-7}})}^{2}}\,{{[C{{l}^{-}}]}^{2}}}=0.777$ subtracting we get $0.177=0.059\log \frac{[{{H}^{+}}]}{{{10}^{-7}}}$ $\Rightarrow$   $\frac{[{{H}^{+}}]}{{{10}^{-7}}}=1000$ $\Rightarrow$   $[{{H}^{+}}]={{10}^{-4}}$$\Rightarrow \,\,\,\,pH=4.$