Consider an electrochemical cell similar to Figure 9.1 with a Pb(s) electrode an
ID: 501005 • Letter: C
Question
Consider an electrochemical cell similar to Figure 9.1 with a Pb(s) electrode and Pb(NO_3)_2(aq) solution in one beaker, and Ag(s) and AgNO_3(aq) in the other. The aqueous solutions both have initial concentrations of 1.0 M. Ag^+(aq) +e^- rightarrow Ag(s) +0.80 V Pb^2+ (aq) + 2e^- rightarrow Pb(s) -0.26 V a. Write down the half reactions, and derive the overall balanced reaction. Also, show the calculation for the standard cell potential. b. As time passes, how will the mass of the Pb electrode change? c. If the Pb^2+(aq) concentration is kept at 1.0 M, but the Ag^+(aq) concentration is increased to a much higher concentration, how will the cell potential change compared to the standard cell potential? d. If both the Pb^2+(aq) and Ag^+(aq) concentrations are changed to 0.01 M, how will the cell potential change, compared to the standard cell potential?Explanation / Answer
a) Pb|Pb(NO3)2||AgNO3|Ag
Half reaction at cathode:
AgNO3(aq) + 1 electron -----> Ag (s) (reduction)
OR
2[AgNO3(aq) + 1 electron -----> Ag (s)] (Multiplying both sides by 2)
Half reaction at anode
Pb(s)----------> Pb(NO3)2 (aq)+ 2 electrons ( oxidation )
Combine both half reactions, (Note that we have multiplied reduction half reaction by 2 so that number of electrons
transferred are same)
Therefore the overall balanced reaction is
2AgNO3(aq) +Pb(s) ------> 2Ag (s) +Pb(NO3)2 (aq)
Calculation of the standard cell potential :-
Half reaction at cathode:
AgNO3(aq) + 1 electron -----> Ag (s) and E(cathode) = +0.80V
OR
2[AgNO3(aq) + 1 electron -----> Ag (s)] (Multiplying both sides by 2) and E(cathode) = +1.60V ------1
Half reaction at anode
Pb(s)----------> Pb(NO3)2 (aq) + 2 electrons and E(anode) = -0.26V------2
Therefore, E(cell) = E(cathode) - E(anode)
= 1.60 - (-0.26)
= 1.60 + 0.26
= +1.86V
b) As the time passes, the mass of Pb electrode will keep on reducing.