The emf of the cell with the cell reaction given below is 0.28V at 25oC. Zn(s)+2H+(aq)→Zn2+(aq,0.1M)+H2(g,1atm
Calculate the pH of the hydrogen electrode
Given: E0Zn2+/Zn=−0.76V E0H+/H2=0V
A
3.47
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B
5.38
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C
8.62
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D
12.95
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Solution
The correct option is C8.62 Standard reduction potential of Zn2+/Zn couple is less than standard reduction potential of hydrogen electrode.
Thus, hydrogen electrode will act as cathode and zinc couple will act as anode.
Here, standard hydrogen electrode is act as reference electrode and its standard reduction potential is zero.
Hence, the standard cell potential is E0cell=E0cathode−E0anode E0cell=0−(−0.76) Eocell=0.76V
The given cell reaction is, Zn(s)+2H+(aq)→Zn2+(aq,0.1M)+H2(g,1atm
According to Nernst equation, Ecell=Eocell−0.05912log[Zn2+]PH2[H+]2 0.28=0.76−0.05912log10(0.1)×1[H+]2 log(0.1)×1[H+]2=2×0.480.0591