For the following voltaic cell, determine the [Cl
-] when
PCl2 = 0.500 atm, [Zn
2+] = 1.77 × 10
-2 M, and
Ecell = 2.250 V. The half-reactions at 25 °C are:
Cl
2(g) + 2 e
-→ 2 Cl
-(aq)
E° = +1.358 V
Zn
2+(aq) + 2 e
-→ Zn(s)
E° = -0.763 V
Zn(s)
Zn
2+(aq)
Cl
-(aq), Cl
2(g)
Pt(s)
◦ 5.48 × 10
-6 M
◦ 2.32 × 10
-3 M
◦ 0.0296 M
◦ 0.0939 M
◦ 0.0352 M