Potentials F2=2.87 Hg(l)=.855 tandard Reduction (ElecX C Chegg Study | Guided Sc

Question

Potentials F2=2.87 Hg(l)=.855 tandard Reduction (ElecX C Chegg Study | Guided Sc X signment/takeCovalentActivity.do?locator assignment-take&takeAssig; ocator assignment-take Use the Refereaces to access important valers if seeded for this questios. What is the calculated value of the cell potential at 29SK for an electrochemical cell with the following reaction, when the F pressure is 5.32-10+ atm, the F concentration is 1.42M, and the Hg2 concentration is 123M? The cell reaction as written above is spostaneous for the concentrations given given Rebry Entire Group mote group ampts remaining Subnit Answe ?? 8 9 6

Explanation / Answer

F2(g) + Hg(l) ------------------ 2 F-(aq) + Hg+2(aq)

according to equation F2 undergoes reduction and Hg undergoes oxidation

at anode oxidation reaction is    Hg(l) ----------------- Hg+2(aq) + 2e-

at cathode reduction reaction is    F2(g) +2e- ------------- 2 F-(aq)

   ----------------------------------------------------------

   Hg(l) + F2(g) ------------------ 2 F-(aq) + Hg+2(aq)

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E0 of F2/2F- = + 2.87V    E0 of Hg+2/Hg = + 0.855V

E0cell = E0cathode - Eo anode

E0cell = 2.87 - 0.855= 2.015V            n=2e-

Ecell = E0cell - 0.0591/n log[ F-]^2[Hg+2]/PF2

Ecell = 2.015 - 0.0591/2 log (1.42)^2 (1.23)/(5.32x10^-4)

Ecell = 2.015 - 0.108

Ecell = 1.907V

Cell potential = 1.907V

The reaction is spontaneous because Ecell is Positive sign.

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