Question

Calculate E∘cell for each of the following balanced redox reactions. Are the reactions spontanous?

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)

Answer 1

A. 2Cu (s) + Mn²⁺ (aq) 2Cu⁺ (aq) + Mn (s)

Half-cell reactions are :

2Cu (s) 2Cu⁺ (aq) + 2e^- (oxidation) E⁰_oxd = -0.520 V

Mn²⁺ (aq) + 2e^- Mn(s) (reduction) E⁰_red = -1.185 V

E⁰_cell = E⁰_oxd + E⁰_red = -0.520 - 1.185 = - 1.705 V

For a reaction to be spontaneous, G⁰ should be negative, We know,

G⁰ = -nFE⁰_cell as, E⁰_cell is negative so G⁰ will be positive, hence the reaction will not be spontaneous.

B. Here the half-cell reactions are:

MnO₂ (s) + 2e- Mn²⁺ (aq) (reduction) E⁰_red = + 1.23 V

Zn (s) Zn²⁺ (aq) + 2e- (oxidation) E⁰_oxd = + 0.7618 V

E^o_Cell = E⁰_red + E^o_oxd = 1.23 + 0.7618 = 1.9918 V

Here E⁰_cell is positive, so G⁰ will be negative, hence the reaction will be spontaneous.

C. Here the half-cell reactions are:

Cl₂ (g) + 2e- 2Cl- (aq) (reductioon) E⁰_red = + 1.36 V

2F- (aq) F₂ (g) + 2e- (oxidation) E⁰_oxd = -2.87 V

E⁰_cell = E^o_red + E^oo_xd = 1.36 - 2.87 = - 1.51 V

Here again as E^o_cell is negative so G⁰ will be positive and thus the reaction will not be spontaneous.