Question

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 1

b. square root of 3 over 3 is the answer to your question!!! I hope I helped!!!!!!!!!! XoXo -Marcey<3! :D

Answer 2

Answer : The correct option is, (B)
`K=6.27* 10^(-91)`

Solution :

The given balanced redox reaction is,

`2Cr^(3+)(aq)+3Sn^(2+)(aq)\rightarrow 2Cr(s)+3Sn^(4+)(aq)`

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

The relation between the equilibrium constant and cell potential :

`\Delta G=-nFE^o_(cell)\\\\\Delta G=-2.303RT\log K`

By equation these two equation we get,

`E^o_(cell)=(0.0592)/(n)* \log K` ........(1)

where,

`E^o_(cell)` = cell potential = -0.89 v

n = number of electrons = 6

K = equilibrium constant

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

`-0.89v=(0.0592)/(6)* \log K`

`\log K=-90.202`

`K=6.28* 10^(-91)`

Therefore, the value of the equilibrium constant is,
`K=6.27* 10^(-91)`