Calculate the potential of hydrogen electrode in contact with a solution whose `pH=10`.
Text Solution
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If pH of solutions is 10 then its `[H^(+)]` ion concentration will be `10^(-10)M`. Let us consider a reduction half cell `H^(+)(10^(-10)M)|H_(2)(1 "atm")|Pt` Electrode process : `2H^(+)(10^(-10)M)+2e^(-)hArrH_(2)(1"atm")(n=2)` `Q=(PH_(2))/([H^(+)]^(2))=(1)/([10^(-10)])=10^(20)` According to to Nernst equation `E_(H^(+)//H_(2))=E_(H^(+)//H_(2))^(0)-(0.059)/(n)"log Q"` `=0-(0.059)/(2)"log10"^(20)=-0.59V`
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Calculate the reduction potential of the following electrodes : a. Pt,H_(2)(4 atm)|H_(2)SO_(4)(0.01M) b. Pt,H_(2)(1 atm)| HCl (0.2 M) c. Calculate the potential of hydrogen electrode in contact with a solution whose i pH=5" "ii. pOH=4
The potential of hydrogen electrode having a pH=10 is
The reduction potential of hydrogen electrode containing a solution of pH=4 is
