Question

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: AgCl(s)+e−→Ag(s)+Cl−(aq). The two cell compartments have [Cl−]= 1.60×10−2 M and [Cl−]= 3.00 M , respectively.

Which electrode is the cathode of the cell?

the compartment with 1.60×10−2 M Cl(aq) is the cathode

Correct

2. What is the standard emf of the cell?

3.What is the cell emf for the concentrations given?

4. For the anode compartment, predict whether [Cl−] will increase, decrease, or stay the same as the cell operates.

For the anode compartment, predict whether  will increase, decrease, or stay the same as the cell operates.

increase

decrease

stay the same

5. For the cathode compartment, predict whether [Cl−] will increase, decrease, or stay the same as the cell operates.

For the cathode compartment, predict whether  will increase, decrease, or stay the same as the cell operates.

increase

decrease

stay the same

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Answer 1

ans)

from baove data that

a)

we have the half reaction:

AgCl (s) + e^- = Ag(s) + Cl^-(aq)

Ag(s) + Cl^-(aq) = AgCl (s) + e^-

According to several sources, and textbooks, the less concentrated ion will be the in the cathode cell to promove the reduction reaction. In this case, we can say that Cl^- = 0.0160 M is the electrode in the cathode cell,

because is reducting from zero to 1, so, it's the cathode.

b)

As we are having a two silver-silver electrode, both reactions have the same emf in each reaction so:

Eº = E (AgCl/Ag) - E (AgCl/Ag) = 0.00 V

the standard emf of the cell= 0.00 V

c)

using the nerst equation as:

E = Eº - RT/nF ln [Cl less] / [Cl more] Where Cl less is the less concentrated ion. R = 8.3144 J/K mol and F = 96485:

E = -8.3144*(298) / 1*96485 ln (0.0160/3)

E = -0.0257 ln (5.33 x10^-3)

E = 0.1345 V

d)

As oxidation is ocurring in the anode according to the half reaction above,

we can say that Cl^- is consumed, therefore it will decrease as the cell operates

.so option B right answer

e)

As reduction is ocurring in the cathode, according to the half reaction above,

we can say that Cl^- is produced, therefore it will increase as the cell operates.

so option A right answer