Consider the following half-reactions: Cl_2(g) + 2 e^- rightarrow 2 Cl (aq) E^=

Question

Consider the following half-reactions: Cl_2(g) + 2 e^- rightarrow 2 Cl (aq) E^= +1.36V Cl_2(g) + e^- rightarrow Ag(s) E^= +0.80V Cu^2+(aq) + 2 e^- rightarrow Cu(s) E^= +0.34V Sn^2+(aq) + 2 e^- rightarrow Sn(s) E^= -0.14V Al^3+(aq) + 3 e^- rightarrow Al(s) E^= -1.66 V Which of the above elements or ions will reduce Cu^2+(aq)? Ag(s) and Sn^2+(aq) Cl^-(aq) and Ag(s) Cl_2(g) and Ag^(aq) Sn(s) and Al(s) Sn^2+(aq) and Al^3+(aq) The SHE electrode has been assigned a standard reduction potential, E^, of 0.00 Volts. Which reaction occurs at this electrode? 2 H_2O(l) + 2 e^- rightarrow H_2(g) + 2 OH^-(aq) O_2(g) + 4 e^- rightarrow 2 O^2-(aq) Hg_2Cl_2(s) + 2 e^- rightarrow 2 Hg(l) + 2 Cl^-(aq) LI^+(aq) + e^- rightarrow Li(s) 2 H^+(aq) + 2 e^- rightarrow H_2(g) Which ONE of the equations below is the correctly balanced equation for the oxidation of cadmium metal, Cd(s), by concentrated nitric acid, HNO_3(aq), producing No_2(g) and Cd^2+(aq)? HNO_3(aq) + Cd(s) rightarrow Cd^2+(aq) + NO_2(g) + OH^-(aq) 2 HNO_3(aq) + Cd(s) rightarrow Cd^2+(aq) + 2 NO_2(g) + 2 OH^-(aq) HNO_3(aq) + Cd(s) + H^+(aq) rightarrow Cd^2+(aq) + NO_2(g) + H_2O(l) 4 HNO_3(aq) + Cd(s) rightarrow Cd^2+(aq) + NO_2(g) + 2H_2O(l) + 2 NO_3^- (aq) HNO_3(aq) + Cd(s) rightarrow Cd^2+(aq) + NO_2(g)

Explanation / Answer

46)

given Cu+2 should be reduced

so

Cu+2 + 2e- ---> Cu (s) Eo = 0.34 V

so

to reduce Cu+2

the other half reaction should have Eo < 0.34

from the given list

the possible reactions are

Sn+2 + 2e- --> Sn (s) Eo = -0.14

Al+3 + 3e- --> Al (s) Eo = -1.66

since

Cu+2 is reduced, the other should be a oxidation reaction

so write the above reactions as oxidation reactions

we get

Sn ---> Sn+2 + 2e-

Al --> Al+3 + 3e-

so

Sn and Al are the elements that will reduce Cu+2

so

the answer is D) Sn and Al

47)

SHE is standard hydrogen electrode

so

at this electrode

the reduction of hydrogen takes place

the reaction is given by

2H+ + 2e- --> H2

so

the answer is E) 2H+ + 2e- --> H2

48)

oxdiation :

Cd ---> Cd+2 + 2e-

reduction :

HN03 --> N02

balance O atoms by using H20

HN03 ---> N02 + H20

now balance H atoms using H+

HN03 + H+ ---> N02 + H20

now balance the charge using electrons

HN03 + H+ + e- ---> N02 + H20

2HN03 + 2H+ + 2e- ---> 2N02 + 2H20

Cd --> Cd+2 + 2e-

the overall reaction is

Cd + 2HN03 + 2H+ --> Cd+2 + 2N02 + 2H20

now add 2 OH- on both sides

Cd + 2HN03 + 2H+ + 2OH- ---> Cd+2 + 2N02 + 2H20 + 2OH-

now H+ + OH- --> H20

so

Cd + 2HN03 + 2H20 ---> Cd+2 + 2N02 + 2H20 + 2OH-

Cd + 2HN03 --> Cd+2 + 2N02 + 2OH-

so

the answer is option B)

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