Avoltaic cell is constructed with an ag/ag* half-cell and a pb/pb^2+ half-cell. which metal will serve as the anode and which as the cathode? justify your answer quantitatively. write a cell reaction for this cell. write the half-reactions and label them as oxidation and reduction. draw a cell diagram for this cell. determine the standard cell potential. if [ag^+] = 0.001 m and [pb2^+] = 0.01 m, determine the cell potential.

Explanation:

From the table of standard reduction potentials we get:

Ag⁺  + e⁻     ⇒    Ag             Eº = 0.80 V

Pb²⁺   + 2 e⁻  ⇒  Pb             Eº = - 0.13 V

Ag⁺   has a greater reduction potential ( more positive) than Pb²⁺, therefore it wil be reduced while Pb will be oxidized.

In the cathode is where reduction occurs, therefore the Ag metal will act as the cathode. Likewise, the lead metal will act as the anode.

The overall reaction is

Pb  ⇒  Pb²⁺ +  2e⁻                        Eº (oxidation) = + 0.13 V        

2 Ag⁺  + 2e⁻     ⇒    2 Ag              Eº (reduction) = +0.80 V

Pb + 2 Ag⁺    ⇒   Pb²⁺  + Ag         Eºcell = 0.13 V + 0.80 V = 0.93 V

Follow the conventions ( oxidation in the right side, junction, reduction ) to represent the cell diagram which in this case is:

Pb(s) ∣ Pb²⁺ (1 M) ∥ Ag+(1 M) ∣ Ag(s)  

For the non standard conditions , [Ag^+] = 0.001 M and [Pb2^+] = 0.01 M , we are going to make use of the Nernst equation:

E cell = Eº cell - 2.303 RT/nF log Q

where,

R = 8.314 J/Kmol

T = 298 K (25ºC)

n = number of electrons transferred

F =   Faraday constant ,96500 coulomb/mol e⁻

Q = reaction quotient = molar concentration Reducing agent/ molar concentration Oxidizing agent

Plugging our values, we have

Q =  [Pb2^+] /  [Ag^+]   =  0.01 M/0.001 = 10

E cell = 0.93 V - ( 2.303 x 8.314 J/Kmol x 298 K ) / ( 2 x 96,500  coul/mol) x log 10

E cell =  0.90 V

Note=  we could have used the form for the Nernst equation:

Ecell = Eº -0.0592/n log Q

which is simpler but valid only at 298 K ( 25 ºC ), the form we used  is more general.

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.