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Reduction half-reaction E∘ (V) Ag+(aq)+e−→Ag(s) 0.80 Cu2+(aq)+2e−→Cu(s) 0.34 Sn4+(aq)+4e−→Sn(s) 0.15 2H+(aq)+2e−→H2(g) 0 Ni2+(aq)+2e−→Ni(s) −0.26 Fe2+(aq)+2e−→Fe(s) −0.45 Zn2+(aq)+2e−→Zn(s) −0.76 Al3+(aq)+3e−→Al(s) −1.66 Mg2+(aq)+2e−→Mg(s) −2.37 1) Use the table of standard reduction potentials given above to calculate the equilibrium constant...

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Chemistry

Reduction half-reactionE∘ (V)
Ag+(aq)+e−→Ag(s)0.80
Cu2+(aq)+2e−→Cu(s)0.34
Sn4+(aq)+4e−→Sn(s)0.15
2H+(aq)+2e−→H2(g)0
Ni2+(aq)+2e−→Ni(s)−0.26
Fe2+(aq)+2e−→Fe(s)−0.45
Zn2+(aq)+2e−→Zn(s)−0.76
Al3+(aq)+3e−→Al(s)−1.66
Mg2+(aq)+2e−→Mg(s)−2.37

1)

Use the table of standard reduction potentials given above tocalculate the equilibrium constant at standard temperature (25 ∘C)for the following reaction: Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s)

2)

Calculate the standard cell potential (E∘) for thereaction

X(s)+Y+(aq)→X+(aq)+Y(s)

if K = 3.80×10−4.

Express your answer to three significant figures and include theappropriate units.

Answer & Explanation Solved by verified expert
3.5 Ratings (504 Votes)
1 FesNi2aqFe2aqNis Fes Fe2aq 2e Eo 045 V Ni2 2e NiEo 026 V Eocell 045 V026 V 019 V no    See Answer
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