Question

What is the E cell for the following

1.) magnesium and zinc

2.) magnesium and copper

3.) copper and zinc

4.) 1 M Cu2+ and 0.01 M Cu2+

Answer 1

1) magnesium and zinc

Zn²⁺(aq) + 2e- → Zn (s) E^o = -0.76 V - reduction

Mg(s)→ Mg²⁺(aq) + 2e- E^o = +2.38 V - oxidation

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Mg(s)+Zn²⁺(aq) → Mg²⁺(aq)+Zn (s)

E^o_cell = E^o_reduction of reaction at cathode+E^o_oxidation of reaction at anode

= -0.76 V + 2.38 V

= + 1.62 V

E^o_cell = + 1.62 V

2)  magnesium and copper

Cu²⁺(aq) + 2e- → Cu(s) E^o = + 0.34 V - reduction

Mg(s)→ Mg²⁺(aq) + 2e- E^o = +2.38 V - oxidation

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Mg(s)+Cu²⁺(aq) → Mg²⁺(aq)+Cu(s)

E^o_cell = E^o_reduction of reaction at cathode + E^o_oxidation of reaction at anode

= +0.34 V + 2.38 V

= + 2.72 V

E^o_cell = + 2.72 V

3)

Cu²⁺(aq) + 2e- → Cu (s) E^o = + 0.34 V - reduction

Zn(s)→ Zn²⁺(aq) + 2e- E^o = + 0.76 V - oxidation

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Zn (s)+Cu²⁺(aq) → Zn ²⁺(aq)+Cu(s)

E^o_cell = E^o_reduction of reaction at cathode + E^o_oxidation of reaction at anode

= +0.34 V + 0.76 V

= + 1.1 V

E^o_cell = + 1.1 V