The standard reduction potential for the half-reaction: is +0.15 V. Consider dat

Question

The standard reduction potential for the half-reaction: is +0.15 V. Consider data from the table of standard reduction potentials for common half-reactions, in your text. For a galvanic cell under standard conditions, which of the following anodic half reactions would produce, at the cathode a spontaneous reduction of Sn^4+ to Sn^2+ but not Sn^2+ to Sn. Ni rightarrow Ni^2+ + 2e^- Fe rightarrow Fe^3+ + 3e^- H_2 rightarrow 2H^+ + 2e^- Pb + SO_4^2- rightarrow PbSO_4 + 2e^- H_2 + 20H^- rightarrow 2H_2O + 2e^- Fe rightarrow Fe^2+ + 2e^-

Explanation / Answer

the standard reduction potentials are

Sn+2 + 2e- ----> Sn ----> Eo = -0.14 V

Ni+2 + 2e- ---> Ni ----> Eo = -0.26 V

Fe+3 + 3e- ---> Fe -----> Eo = -0.04 V

2H+ + 2e- ---> H2 ------> Eo = 0 V

Fe+2 + 2e- ---> Fe ----> Eo = -0.44 V

PbS04 + 2e- ----> Pb + S042- ----> Eo = -0.36 V

2H20 + 2e- ---> H2 + 2OH- ----> Eo = -0.83 V

now

we know that

for a cell reaction to be spontaneous

Eo cell > 0

now

Eo cell = Eo cathode - Eo anode

1) in this case the given cathode reaction is

Sn+4 + 2e- ---> Sn+2 ----> Eo = 0.15

now

Eo cell = Eo cathode - Eo anode > 0

0.15 - Eo anode > 0

Eo anode < 0.15

so

to make the cathodic reaction of Sn+4 ---> Sn+2 spontaneous

the anode shoule have Eo < 0.15 V

2)

now

the cathodic reaction is

Sn+2 + 2e- ----> Sn -----> Eo = -0.14 V

now

in this case , given that

the reaction should not be spontaneous

so

Eo cell = Eo cathode - Eo anode < 0

-0.14 - Eo anode < 0

-Eo anode < 0.14

Eo anode > -0.14

now

from the given list

the anodes that satisfy the above condition are

H2 ---> 2H+ + 2e-

Fe ---> Fe+3 + 3e-

So

the given conditions are true for two of the given anodes

H2 ---> 2H+ + 2e-

Fe ---> Fe+3 + 3e-

the rest does not satisfy the given conditions

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