1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The emf of the cell with the cell reaction given below is 0.28 V at 25oC. Zn(s) + 2H+(aq)→Zn2+(aq, 0.1 M) + H2(g, 1 atm Calculate the pH of the hydrogen electrode Given: E0Zn2+/Zn = −0.76 V E0H+/H2 = 0 V

A
3.47
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5.38
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
8.62
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
12.95
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 8.62Standard 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 = 0.76 V The given cell reaction is, Zn(s) + 2H+(aq)→Zn2+(aq, 0.1 M) + H2(g, 1 atm 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 log 0.1 − log [H+]2 = 16.24 − log [H+]2=16.24−log 0.1 −2 log [H+]=16.24+1 2 pH = 16.2436+1 ∵−log [H+]=pH pH = 17.242 pH= 8.62

Suggest Corrections
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program