please explain this to me?
2017-12-11 2:47 am
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.857 M and [Fe2 ] = 0.0160 M. Standard reduction potentials can be found here.
回答 (2)
2017-12-11 3:14 am
http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html
Refer to the above webpage for standard reduction potentials.
Different answer would be obtained for different source of standard reduction potentials used.
Mg²⁺(aq) + 2e⁻ → Mg(s) .… E° = -2.38 V
Fe²⁺(aq) + 2e⁻ → Fe(s) …. E° = -0.41 V
Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s) …. E°(cell) = (-0.41) - (-2.38) V = +1.97 V
E(cell)
= E°(cell) - (RT/zF) ln([Mg²⁺]/[ Fe²⁺])
= +1.97 - {8.314 × (273.16 + 25.00) / 2 × 96500} ln (0.815/0.0180) V
= +1.92 V
Refer to the above webpage for standard reduction potentials.
Different answer would be obtained for different source of standard reduction potentials used.
Mg²⁺(aq) + 2e⁻ → Mg(s) .… E° = -2.38 V
Fe²⁺(aq) + 2e⁻ → Fe(s) …. E° = -0.41 V
Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s) …. E°(cell) = (-0.41) - (-2.38) V = +1.97 V
E(cell)
= E°(cell) - (RT/zF) ln([Mg²⁺]/[ Fe²⁺])
= +1.97 - {8.314 × (273.16 + 25.00) / 2 × 96500} ln (0.815/0.0180) V
= +1.92 V
2017-12-11 3:13 am
Add them up then divide it by the most commonly reactive metal. Then it's system of equations from there
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