Question

Calculate ΔE° for the following equation (with solutions):

3Br₂ + 2Cr³⁺ + 7H₂O = Cr₂O^2-₇ + 14H⁺ + 6Br^-

Answer 1

3Br₂ + 2Cr³⁺ + 7H₂O Cr₂O₇^2- + 14H⁺ + 6Br^-

Reduction half: 3Br₂ + 6e^-   6Br^- : E^o_red = +1.07 V

Oxidation half:   2Cr³⁺ + 7H₂O   Cr₂O₇^2- + 14H⁺ + 6e^- : E^o_oxid = -1.33 V

E^o_cell = E^o_red + E^o_oxid = 1.07 V - 1.33 V = - 0.26 V

If the E^o_cell value had been positive it would have meant that reaction is spontaneous. However, a negative E°_cell indicates that the reaction will proceed spontaneously in the opposite direction. This is because reversing the direction of a reaction changes the sign of it reduction potential. So, if the direction of the reaction given in the question is reversed then its reduction potential will be changed to + 0.26 V. This is a positive value and hence in this direction the reaction would be spontaneous.

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