What is the value of the equilibrium constant for this redox reaction?
2Cr^3+(aq) + 3Sn^2+ (aq) 2Cr(s) + 3Sn^4+(aq) E = -0.89 v

Question

What is the value of the equilibrium constant for this redox reaction?
2Cr^3+(aq) + 3Sn^2+ (aq) 2Cr(s) + 3Sn^4+(aq) E = -0.89 v

anonymous · Answer

K = 5.13 x 10-93




B.
K = 6.27 x 10-91 




C.
K = 7.92 x 10-46 




D.
K = 8.56 x 10-31 




E.
K = 9.25 x 10-16

chmvijay · Answer

E = 0.0591 /n Log Kc
can u find Kc 
E= -0.89
and n = no of electrons involved in the reaction

chmvijay · Answer

we dont give direct answer here @david_crider  we give u hint or guide u have to solve them ur self now i have give u the formula u just put them in that formula and find out

anonymous · Answer

Im in summer school for a reason haha i dont know how to apply the formula, just trying to get through this

anonymous · Answer

@chmvijay