Calculate Eº cell for each of the following balanc...

Question

Calculate Eº cell for each of the following balanced redox reactions.

Determine whether the reaction is spontaneous as written.

a. 2Cu(s) +Mn²⁺(aq) → 2Cu⁺(aq) + Mn(s)
b. MnO₂(s) + 4H⁺(aq) + Zn(s) → Mn²⁺(aq) + 2H₂O(l) + Zn²⁺(aq)
c. Cl₂(g) + 2F⁻(aq) → F₂(g) + 2Cl⁻(aq)

Jules Bruno · Accepted Answer

E°cell=E°cathode-E°anode Recall:Lose GainElectron ElectronsOxidation Reductioncathode → reduction → oxidation number decreasesanode → oxidation → oxidation number increasesFor ions: charge of the ion = oxidation numberFor neutral atoms/compound: oxidation number = 0Also recall:E°cell > 0 → spontaneous in the forward directionE°cell < 0 → nonspontaneous in the forward direction[readmore]a. 2Cu(s) + Mn2+(aq) → 2Cu+(aq) + Mn(s)Half-cell reactions:2Cu(s) → 2Cu+(aq)charge changed from 0 to +1Cu = 0, Cu+ = +1oxidation number increased → lost electrons → oxidized → anodeMn2+(aq)→ Mn(s)charge changed from +2 to 0Mn+2 = +2, Mn = 0oxidation number decreased → gained electrons → reduced → cathodeCalculate E°cell.Cu+(aq) + e- → Cu(s) E°cell = 0.52 V anodeMn2+(aq) + 2 e- → Mn(s) E°cell = -1.18 V cathode E°cell=E°cathode-E°anodeE°cell=(-1.18 V )-0.52 V E°cell = -1.7 V → E°cell < 0 → nonspontaneousb. MnO2(s) + 4H+(aq) + Zn(s) → Mn2+(aq) + 2H2O(l) + Zn2+(aq)Half-cell reactions:Zn(s)→ Zn2+(aq)charge changed from 0 to +2Zn = 0, Zn2+ = +2oxidation number increased → lost electrons → oxidized → anode MnO2(s) + 4H+(aq) → Mn2+(aq) + 2H2O(l) charge changed from +4 to +2sum of oxidation states neutral cpd. = 0If O.S. of Mn = xx + 2(O.S. of O) = 0x + 2(-2) = 0x = + 4 Mn+2 = +2oxidation number decreased → gained electrons → reduced → cathodeCalculate E°cell.MnO2(s) + 4 H+(aq) + 2 e- → Mn2+(aq) + 2 H2O(l) E°cell = 1.21 V cathodeZn2+(aq) + 2 e- → Zn(s) E°cell = -0.76 V anode E°cell=E°cathode-E°anodeE°cell=(1.21 V )--0.76 V E°cell = 1.97 V → E°cell > 0 → spontaneousc. Cl2(g) + 2F−(aq) → F2(g) + 2Cl−(aq)Half-cell reactions:Cl2(g)→ 2Cl−(aq)charge changed from 0 to -1Cl2 = 0, Cl- = -1oxidation number decreased → gained electrons → reduced → cathode2F−(aq) → F2(g) charge changed from -1 to 0F- = -1, F2 = 0, oxidation number increased → lost electrons → oxidized → anodeCalculate E°cell.F2(g) + 2 e- →2 F-(aq) E°cell = 2.87 V anodeCl2(g) + 2 e- →2 Cl-(aq) E°cell = 1.36 V cathode E°cell=E°cathode-E°anodeE°cell=(1.36 V )-2.87 V E°cell = -1.51 V → E°cell < 0 → nonspontaneous

Problem: Calculate Eº cell for each of the following balanced redox reactions.Determine whether the reaction is spontaneous as written.a. 2Cu(s) +Mn2+(aq) → 2Cu+(aq) + Mn(s)b. MnO2(s) + 4H+(aq) + Zn(s) → Mn2+(aq) + 2H2O(l) + Zn2+(aq)c. Cl2(g) + 2F−(aq) → F2(g) + 2Cl−(aq)

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Problem: Calculate Eº cell for each of the following balanced redox reactions.Determine whether the reaction is spontaneous as written.a. 2Cu(s) +Mn2+(aq) → 2Cu+(aq) + Mn(s)b. MnO2(s) + 4H+(aq) + Zn(s) → Mn2+(aq) + 2H2O(l) + Zn2+(aq)c. Cl2(g) + 2F−(aq) → F2(g) + 2Cl−(aq)

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