The standard reduction potential for the half-reaction Sn^4+ + 2e^- rightarrow S

Question

The standard reduction potential for the half-reaction Sn^4+ + 2e^- rightarrow Sn^2+ 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 cathodic half reactions would produce at the anode, a spontaneous oxidation of Sn to Sn^2+ but not Sn^2+ to Sn^4+. Sn^4+ + 2e^- rightarrow Sn^2+ Pb^2+ + 2e^- rightarrow Pb 2H^+ + 2e^- rightarrow H_2 Fe^2+ + 2e^- rightarrow Fe PbSO_4 + 2e^- rightarrow Pb + SO_4^2- 2H_2O + 2e^- rightarrow H_2 + 2OH^-

Explanation / Answer

Sn^4+ + 2e- Sn^2+ E0 = 0.15 v

2 H+ + 2 e H2(g) E0 = 0.0000 v

Pb2+ + 2 e Pb(s) E0 = 0.126 v

Sn2+ + 2 e Sn(s) E0=    0.13 v

PbSO4(s) + 2 e Pb(s) + SO42 E0 = 0.3588 v

Fe2+ + 2 e Fe(s) E0 = 0.44 v

2H2O + 2 e H2(g) + 2 OH E0 = 0.8277 v

in a galvanic cell ,

the electrode with higher reduction potential acts as cathode involves in reduction.

similarly, the electrode with lower reduction potential acts as anode involves in oxidation.

so that,   Sn2+ + 2 e Sn(s) E0=    0.13 v

if the reduction potential is greater than -0.13 v , that can make the sn to involve oxidation to sn^2+ , and the electrode

which has potential less than 0.15 v donot oxidise the Sn^2+ into Sn^4+.

so that ,

answer :

Sn^4+ + 2e- Sn^2+ E0 = 0.15 v

2 H+ + 2 e H2(g) E0 = 0.0000 v

Pb2+ + 2 e Pb(s) E0 = 0.126 v

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