Question

Ten millilters of a solution containing 0.0835 M Cr⁺³ and 0.119 M Cr⁺² is added to 25.0 mL of a solution containing 0.0361 M V⁺³ and 0.0904 M V²⁺. Does a reaction occur, if so, what direction?

Answer 1

the reductions potentials:

Cr3+ + e− ⇌ Cr2+ −0.42

V3+ + e− ⇌ V2+ −0.26

Recalcualte concentrations:

[Cr+3] = M1V1/(V1+V2) = 0.0835*10 / (10+25) = 0.023857

[Cr+2] = M1V1/(V1+V2) = 0.119*10 / (10+25) = 0.034

[V+3] = M2V2 /(V1+V2) = 0.0361*25/(25+10) = 0.0257

[V+2] = M2V2 /(V1+V2) = 0.0904*25/(25+10) = 0.064571

assume this goes in forward direction...

so

Cr must be oxidized:

Cr2+ ⇌ Cr3+ + e− +0.42

V3+ + e− ⇌ V2+ −0.26

Q = [Cr3+][V+2] / [Cr+2][V+3]

Q = (0.023857)(0.064571) / (0.034*0.0257) = 1.7629

E°cell = -0.26 + 0.42 = 0.16 V

so

Ecell = E°cell- 0.0592/n*log(Q)

Ecell = 0.16 - 0.0592/1*log(1.7629)

Ecell = 0.14542 V

so this must go in the forward direction

the reaciton is going to occur:

Cr2+ ⇌ Cr3+ + e−  +0.42

V3+ + e− ⇌ V2+ −0.26

Cr2+ + V3+ -- >Cr3+ +  V2+