You are given the following cell at 25 degree C: Pb(s) | Pb^2+(aq, 0.60M) || Sn^

Question

You are given the following cell at 25 degree C: Pb(s) | Pb^2+(aq, 0.60M) || Sn^2+ (aq, 0.1 M), Sn^4+ (aq, 0.15 M) |Pt(s) Calculate E_cell and compare the spontaneity of the cell as written to the cell under standard conditions. For #1 above, describe how you could determine if the reaction is driven by enthalpy or entropy. Ohm's law states that the electric potential (E) across any two points is the product of the current (I) and the resistance (R) between those two points (E = IR, 1V = 1A * 1 Ohm). We want you to think about any possible danger to you from the current you generate from your electrochemical cells in this lab. A human can feel a current (I) as small as 5 mA. Dry human skin offers a resistance (R) of about 100,000 Ohms. Given this resistance, would you expect to be able to feel a current if you accidentally touched both of the electrodes in any of the cells in this lab simultaneously? Wet skin has a much lower resistance, around 500 Ohms. Would you feel the current with wet skin?

Explanation / Answer

1) The cell is given as

Pb (s) Pb2+ (aq, 0.60 M)Sn2+ (aq, 0.10 M), Sn4+ (aq, 0.15 M)Pt (s)

In a cell representation, the oxidation half is written on the left followed by the reduction half. Therefore, the half reactions are:

Oxidation: Pb (s) ------> Pb2+ (aq) + 2 e- ; E0ox = +0.13 V (the value is obtained by reversing the standard reduction potential value)

Reduction: Sn4+ (aq) + 2 e- ------> Sn2+ (aq); E0red = +0.15 V

The cell reaction is

Sn4+ + Pb (s) ------> Sn2+ (aq) + Pb2+ (aq)

The standard cell potential is given by E0cell = E0red + E0ox = (+0.15 V) + (+0.13 V) = +0.28 V

The given cell is under non-standard conditions (temperature is kept at 25C, i.e, T= 298 K). The cell potential is given by

Ecell = E0cell + 2.303*(R*T/nF)*log [Sn2+][Pb2+]/[Sn4+] (F = 1 Faraday of electricity = 96485 J/V)

Plug in values and obtain

Ecell = (+0.28 V) + [2.303*(8.314 J/mol.K)*(298 K)/(2 mole)*(96485 J/V)]*log (0.10).(0.60)/(0.15)

= (+0.28 V) + (0.02957 V)*log (0.4) = (+0.28 V) + (0.02957 V)*(-0.40) = 0.268 V 0.27 V (ans).

Since Ecell and E0cell are both positive, the reaction is spontaneous (ans).

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