Part A The following cell has a potential of 0.45 V at 25°C. Pt(s) | H2(1 atm) |
ID: 524189 • Letter: P
Question
Part A
The following cell has a potential of 0.45 V at 25°C.
Pt(s) | H2(1 atm) | H+(? M) || Cl–(1 M) | Hg2Cl2(s) | Hg(l)
The standard half-cell potential for the half-reaction Hg2Cl2(s) + 2 e– 2 Hg(l) + 2 Cl–(aq) is 0.28 V. What is the pH in the anode compartment? The following cell has a potential of 0.45 V at 25°C. Pt(s) | H2(1 atm) | H+(? M) || Cl–(1 M) | Hg2Cl2(s) | Hg(l) The standard half-cell potential for the half-reaction Hg2Cl2(s) + 2 e– 2 Hg(l) + 2 Cl–(aq) is 0.28 V. What is the pH in the anode compartment?
12.3
4.7
2.9
7.6
Explanation / Answer
Applying Nernst equation,
Ecell = E0cell - (0.0591/n) * LogQ
0.45 = 0.28 - (0.0591/2)*LogQ
LogQ = - 5.75
Q = 10-5.75
Q = 1.78 * 10-6
Cell reaction is,
H2 (g) + Hg2Cl2 (s) ---------> 2 H+ (aq.) + 2Hg(l) + 2 Cl- (aq.)
So, Q = [Cl-]2[H+]2
[Cl-]2[H+]2 = 1.78 * 10-6
(1)[H+]2 = 1.78 * 10-6
[H+] = 1.33 * 10-3 M
But,
pH = - Log[H+]
pH = - Log(1.33*10-3)
pH = 2.9