Question

Part III (Plus answers to questions posted in the laboratory manual):

i) What are Q, [Ag+(aq)anode], and [Ag+(aq)cathode] when Emeasured becomes 0 mV for a concentration cell prepared from half cells having initial concentrations of 1.0 and 0.01 M. respectively?

ii) Assuming that both solution volumes were 10.0 mL for the concentration cell described in the above question, how much electricity (in coulombs) flowed from the time the circuit was completed until equilibrium was established?

Answer 1

Solution:

(i) The half-cell reactions involves the following:

Ag (s) -----> Ag⁺(aq) + e                       (Anode)

Ag⁺(aq) + e --------> Ag(s)                    (Cathode)

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Ag (s) + Ag⁺(aq) ---------> Ag⁺(aq) + Ag(s)       (Overall cell reaction)

                                

We know for concentration cell:

E = 0.0591/n [log (C2/C1)]; C1 and C2 are the concentration at cathode and anode respectively.

Now when, E measured becomes 0 mV for a concentration cell then we have,

Or, 0 = 0.0591/n (logQ); where, Q = C2/C1

Or, 0 = 0.0591/1x[log (Q]

Or, 0 = 0.0591x logQ

Or, logQ = 0

Or, Q = 10⁰ = 1

(ii) Now, C1 = 0.01M= 0.01 moles Ag⁺ ions/L

Therefore, 0.01L (=10 mL) of solution = 0.01x0.01 = 1x10^-4 moles of Ag⁺ ions

1 mole of Ag+ ions carries = 96500 C

1x10^-4 moles of Ag⁺ ions carries = 96500x1x10^-4 C

                                                    = 9.65 C electricity.