For the following voltaic cell, determine the [CI] when PC12=0.500 atm, [Zn2+]= 1.77 × 102 M, and Ecell-2.250 V. The half-cell potantials at 25°C are: 20p (E°+1.358 V (Cl2(g), Cl(aq)) and E°= -0.763 V (Zn2+(aq), Zn(s)) Zn(s) | Zn2+(aq) || Cl(aq), Cl2(g) | Pt(s) Faraday Constant (F) = 96485 C/mol