Pb+2 +2e- ---> Pb(s) Eo= -0.13V Ag+ + le- ---> Ag(s) Eo= +.80V Pb(s)| Pb2+ (1M)

ID: 546008 • Letter: P

Question

Pb+2 +2e- ---> Pb(s) Eo= -0.13V

Ag+ + le- ---> Ag(s) Eo= +.80V

Pb(s)| Pb2+ (1M) || Ag+ (1M) | Ag(s)

Determine which of the following statements about the cell shown are

True or False.

Cations move to the silver half-cell.

The standard cell potential, o, equals 1.73 V.

As the reaction proceeds, the concentration of the lead ions decreases.

The cell, as represented by the line notation, is a galvanic cell.

The lead half-cell is the anode.

The mass of the lead electrode is decreasing.

Electrons are spontaneously produced in the silver half-cell.

The silver electrode is the anode.

The half reactions are

Pb(s) -------> Pb+2 (aq) + 2e E0 = -0.13V Anode

2Ag+(aq) + 2e -------> 2Ag(s) E0 = +0.80 V Cathode

Overall reaction is

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

a) Cations move to the silver half cell : True

While reaction proceeding Ag+ decrease in Silver half cell, so cation moves to the silver half cell

b) Standard cell potential E0(cell) = +1.73 V : False

E0(cell) = E0(cathode ) - E0 ( anode )

= +0.80V - (-0.13V)

= +0.93V

c) As the reaction proceed concentration of lead ions decreases : False

concentration of Pb+2 increase

d) The cell represented by line notatio is galvanic cell : True

e) The lead half cell is anode :True

f) Mass of lead electrode is decreasing : True

g) Electrons are spontaneously produced in the silve half cell : False

Electrons are produced in lead half cell

h) The silver electrode is anode: False

Silver electrode is cathod

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