Consider the following electrochemical cell: Calculate the standard cell potenti

Question

Consider the following electrochemical cell: Calculate the standard cell potential of the cell as written (right - left). Write the balanced half reactions and the overall spontaneous cell reaction. Half reaction on the right (reduction): Half reaction on the left (oxidation): Overall spontaneous cell reaction (balanced): Write the expression for the equilibrium constant K_a for the spontaneous cell reaction Calculate K_aq for the spontaneous cell reaction Calculate the cell potential, delta E, with the following conditions: In a separate experiment, a gas mixture of nitrogen and H_2 = 0.20 atm is bubbled through a solution initially with [Fe^3+] = 0.005 M and no Fe^2+. The nitrogen is inert and does not react. The solution has added HCI, [HCI] = 0.5 so that the H^+ concentration can be considered constant at 0.5M. Calculate the remaining Fe^3+ concentration, [Fe^3+] once the system would reach equilibrium. T = 298K

Explanation / Answer

(a)

E0cell = E0cathode - E0anode

In this cell, H2|HCl is the anode.

So,

E0cell = 0.77-0 = 0.77 V

(b)

Reduction half cell reaction: Fe3+ + e- ---> Fe2+

Oxidation half cell reaction: H2 ---> 2H+ + 2e-

Overall reaction: 2Fe3+ + H2 ---> 2Fe2+ + 2H+

(c)

Keq = ([Fe2+]2*[H+]2)/([Fe3+]2*pH2)

(d)

Use equation:

E0cell = (0.0591/n)*log K

Putting values:

0.77 = (0.0591/2)*log K

Solving we get:

Keq = 1026 approximately

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