A galvanic cell is made up of a copper electrode in a 1.0 M copper (II) sulfate

Question

A galvanic cell is made up of a copper electrode in a 1.0 M copper (II) sulfate solution, a silver electrode in a 1.0 M silver nitrate solution, and a salt bridge connecting them at 22 degree C. On the diagram below, label the two electrode as(I) anode or cathode, and(2) copper or silver Label the two solution as copper (II) sulfate of silver nitrate, and then show the flow direction of electron through the wire. Write the spontaneous reaction for the Galvanic cell. Calculate the potential for the Galvanic cell. 6.0 M ammonia is added to the copper (II) sulfate solution until the Cu^2+ ion is essentially all converted to the tetraamminecopper(ll) complex ion. At this point the potential of the cell goes up to 0.92 V. Find the residual concentration of Cu^2+ ion in the cell.

Explanation / Answer

a)

anode --> oxdiation occurs here (electorns are obtained)

cathode -> reduciton occurs here(electrons are delivered here)

Cu2+ + 2 e Cu(s) +0.337

Ag2+ + e Ag+ +1.98

the least reduction potential is tha tof Cu, so it will be oxidized, and AG reduced

the electrons flow from solid copper to Ag+ solution

b)

the spontanous reaction

2Ag+(aq) + Cu(s) --> 2Ag(s) + Cu2+(aq)

c)

the potential

E°cell = Ereducction - Eoxidation = 1.98- 0.337 = 1.643 V

d)

residual concentration:

Ecell = E°cell - 0.0592/n*log(Cu2+ / Ag^2)

0.92= 1.643 - 0.0592/2*log(Cu2+ / 1^2)

solve for [Cu+2]

(0.92-1.643) *2/(-0.0592) = log(Cu2+)

-24.425 = log([Cu2+])

[Cu+2] = 10^-24.425 = 3.75837*10^-25

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