What is the value of the equilibrium constant for this redox reaction? 2cr3+(aq) + 3sn2+ (aq) 2cr(s) + 3sn4+(aq) e = -0.89 v a. 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

Answer : The correct option is, (B)

Solution :

The given balanced redox reaction is,

Now we have to calculate the equilibrium constant for the redox reaction.

The relation between the equilibrium constant and cell potential :

By equation these two equation we get,

      ........(1)

where,

= cell potential = -0.89 v

n = number of electrons = 6

K = equilibrium constant

Now put all the given values in equation (1), we get

Therefore, the value of the equilibrium constant is,