Which of the following is the strongest reducing agent? Sn^2+(aq)+rightarrow Sn(

Question

Which of the following is the strongest reducing agent? Sn^2+(aq)+rightarrow Sn(s) E^degree = -0.14 V Sn^4+(aq)+2cdegree rightarrow Sn^2 + (aq) E^degree = 0.15 V Cr^3+(aq)+3edegree rightarrow Cr(s) Edegree = -0.73 V A)Cr(s) B) Sn(s) C) Sn^2+(aq) D) Sn^4+(aq) E) Cr^3+(aq) 17) What clement is being reduced in the following redox reaction? MnO_4-(aq) + H_2c_2O_4(aq) rightarrow Mn^2+(aq) + CO_2(8) A) Mn B) H C)C D) O 18) Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in fornt of H_2C_2O_4 = 1,H_2O = 1 B) H2C_2O_4 = 5, H_2O = 1 C) H_2C_2O_4 = 3, H_2O = 2 D) H_2C_2O_4 = 5, H_2O = 8 E) H_2C_2O_4 = 1, H_2O = 4 19) Use the tabulated half-cell potentials below to calculate the equilibrium constant (K) for the following balanced redox reaction at 25degree C. 2Al(x) + 3 Mg^2+(aq) rightarrow 2 Al^3+(aq) + 3 Mg(s) Al^3+(aq) + 3e- rightarrow Al(s) Edegree = -1.66 V Mg^2+(aq) + 2e- rightarrow Mg(s) Edegree = -2.37 V A) 1.1 times 10^-72 B)4.6 times 10^31 C) 8.9 times 10^73 D) 1.0 tiems 10^24 E) 1.1 tiems 10^72

Explanation / Answer

16)

we know that

lesser the reduction potential , stronger the reducing agent

in this case

the half reaction with least reduction potential is

Cr+3 + 3e- ---> Cr

now reducing agent undergoes oxidation

so

Cr ---> Cr+3 + 3e-

so

Cr is the strongest reducing agent

the answer is A) Cr (s)

17)

write the oxidation number on each side

Mn :

reactants : +7

products : +2

O :

reactants : -2

products : -2

H :

reactants : +1

products : +1

C :

reactants : +3

products : +4

we know that

reduction is decrease in oxidation number

so

Mn is the element being reduced

so

the answer is A) Mn

18)

Mn04- --> Mn+2

balance O atoms using H20

Mn04- ---> Mn+2 + 4H20

balance H atoms using H+

Mn04- + 8H+ --> Mn+2 + 4H20

balance charge using electrons

Mn04- + 8H+ + 5e- --> Mn+2 + 4H20

now

H2C204 --> C02

H2C204 ---> 2 C02

H2C204 ---> 2 C02 + 2H+

H2C204 --> 2 C02 + 2H+ + 2e-

5H2C204 --> 10 C02 + 10H+ + 10e-

2Mn04- + 16H+ + 10e- --> 2Mn+2 + 8H20

so

the overall reaction is

2Mn04- + 5H2C204 + 6H+ --> 10 C02 + 2Mn+2 + 8H20

the answer is

D) H2C204 = 5 , H20 = 8

19)

we know that

oxidation takes place at anode

and

reduction takes place at cathode

so

oxidation : anode

2 Al ---> 2 Al+3 + 6e-

reduction : cathode

3 Mg+2 + 6e- --> 3 Mg

we knoow that

Eo cell = Eo cathode - Eo anode

Eo cell = Eo Mg+2/Mg - Eo Al+3/Al

Eo cell = -2.37 + 1.66

Eo cell = -0.71

now

we know that

dGo = -nFEo = -RTlnKeq

so

nFEo = RTlnKeq

here

n = 6 as six electrons are transferred

so

6 x 96485 x -0.71 = 8.314 x 298 x lnKeq

Keq = 8.93 x 10-73

so

the answer is C) 8.9 x 10-73

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