Question

A student makes a voltaic cell with a Ag electrode in 1.0 M AgNO₃ solution and a Pb electrode in a 1.0 M Pb(NO₃)₂ solution.

a. Identify the cathode and write the half reaction.

b. Identify the anode and write the half reaction.

c. Write the overall reaction and calculate E^o_cell for the voltaic cell.

d. What is E_cell if the concentrations of the ions in solution are [Ag⁺] = 0.045 M and [Pb²⁺] = 0.36 M? Temperature = 298.15 K.

Answer 1

The two half reactions involved in the redox reaction are

Ag⁺ (aq) + e^- --------> Ag (s); E⁰₁ = 0.80 V

Pb²⁺ (aq) + 2 e^- --------> Pb (s); E⁰₂ = -0.13 V

The more positive the value of the standard reduction potential, the more is the tendency of the element to get reduced. Consequently, Ag⁺ will get reduced and Pb will get oxidized.

a) Reduction occurs at the cathode. Since Ag⁺ has a greater tendency to get reduced, Ag⁺ is reduced to Ag at the cathode. The cathode reaction is

Cathode: Ag⁺ (aq) + e^- --------> Ag (s); E⁰_cat = 0.80 V

b) Oxidation occurs at the anode. Pb is oxidized at the anode to Pb²⁺. The anode reaction is

Anode: Pb (s) -------> Pb²⁺ (aq) + 2 e^-; E⁰_anode = 0.13 V (reversing the reaction changes the sign of E⁰).

c) The overall cell reaction is

2 Ag⁺ (aq) + Pb (s) -------> 2 Ag (s) + Pb²⁺ (aq)

E⁰_cell = E⁰_cat + E⁰_anode = (0.80 V) + (0.13 V) = 0.93 V.

d) Write down the Nernst equation as

E_cell = E⁰_cell – R*T/nf*ln [Pb²⁺]/[Ag⁺]² where R = 8.314 J/mol.K is the gas constant; T = 298.15 K; n = number of moles of electrons involved in the redox reaction = 2 moles and F = 96500 J/V. Plug in values and write,

E_cell = 0.93 V – (8.314 J/mol.K)*(298.15 K)/(2 mole).(96500 J/V)*ln (0.36)/(0.045)² = 0.93 V – (0.01284 V)*ln (177.7778) = 0.93 V – (0.01284 V)*(5.1805) = 0.93 V – 0.0665 V = 0.8635 V ≈ 0.86 V (ans).