Answer:
pH ˜ 1.39
Explanation:
solution
given the cell equation as follows:
Pt | H2 (1 atm); H+ (? M) || Hg2Cl2(s); Cl- (1 M) | Hg
half of the reactions are
2H+ + 2 e- --> H2 (E0 = 0.00 V )
Hg2Cl2 + 2 e- --> 2Hg + 2Cl- (E0 = 0.268 V)
E0 cell = E0 cathode -E0 anode
= 0.268 V -0.00V ( the more of positive is cathode while the smaller is anode )
= 0.268V
knowing that Ecell = E0 cell - 0.0592 V/n) log [oxd/red]
0.35 V = 0.268V -(0.0592V/2) log [H+ ]2 /[Cl- ]
0.35V-0.268V =- 0.0296V*2 log [H+ ]/[1.0]
0.082V = -0.0592V log [H+ ]
so 0.082V /0.0592V = - log [H+ ]
1.385 = - log [H+ ]
1.385 = pH (since pH = - log [H+ ] )
In conclusion, pH ˜ 1.39