The standard reduction potentials for the following half-reactions are given as:

Question

The standard reduction potentials for the following half-reactions are given as: Cr^+3e- rightarrow Cr Edegree = -0.73V Br_2+2e^- rightarrow 2Br Edegree = +1.09 V What is the Edegree for this galvanic cell? a. 1.82 V b. 0.36 V c. 4.75 V d. 1.79 V e. 0.72 V 28. In the cell shown, which reaction occurs at the cathode? Cu(s)|Cu^2+(aq)||Ag^+(aq)|Ag(s) a. Cu^2+(aq) + 2e^- rightarrow Cu(s) b. Ag(s) rightarrow Ag^+(aq) + e^- c. Ag^+(aq) + e' rightarrow Ag(s) d. Cu(s) rightarrow Cu^2+(aq) + 2 e^- e. None of the above 29. Consider the following standard reduction potentials in acid solution: Cr^3+ + 3e^- rightarrow Cr Edegree = -0.74 V Co^2+ + 2e^- rightarrow Co Edegree = -0.28 V MnO_4^- + 8H^+ + 5e^- rightarrow Mn^2+ + 4H_2O Edegree= +1.51 V The strongest oxidizing agent listed above is a. Cr^3+ b. Cr c. Mn^2+ d. Co^2+ e. MnO_4^- 30. What is the value of DeltaGdegree at 298K for the following reaction? Pb(s) + 2 H^+(aq) Pb^2+(aq) + H_2(g) Edegree = 0.13 V a. 12 kJ b. -25000 kJ c. 25 kJ d. -25 kJ e. 39 kJ 31. What mass in grams of copper will be deposited from a solution of Cu^2+ by a current of 2.50 A in 2.00 hr? a. 11.9 b. 5.93 c. 23.7 d. 1.65 e. 5.00

Explanation / Answer

27)

we know that

electrode with higher reduction potential is cathode

so

Br2 is cathode and Cr is anode

now

Eo cell = Eo cathode - Eo anode

so

Eo cell = 1.09 + 0.73

Eo cell = 1.82 V

answer is option a) 1.82 V

28)

in the cell notation

cathode is represented on the right side

so

in this case

cathode is Ag+ +e- ---> Ag (s)

29)

we know that

higher the reduction potential , better the oxidizing agent

from the list given

MnO4- has the highest reduction potential

so

MnO4- is the strongesr oxidizing agent

so

answer is e) MnO4-

30)

we know that

dGo = -nFEo

consider

Pb ---> Pb+2 + 2e-

here

n = 2 as two electrons are transferred

so

dGo = -2 x 96485 x 0.13

dGo = -25 x 1000

dGo = -25 kJ

so

the answer is d) -25 kJ

31)

we know that

according to faradays first law of electrolysis

mass deposited is

m = I x t x M / F x z

given

current (I) = 2.5

time (t) = 2 hr = 2 x 60 x 60 = 7200 s

molar mass (M) = 63.5

F= faradays constant = 96485

z = charge = 2

so

m = 2.5 x 7200 x 63.546 / 96485 x 2

m = 5.927

so

mass depostied is 5.93 grams

so

the answer is b) 5.93

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