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

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Answer 2

Answer : The correct option is, (B) $K=6.27\times 10^{-91}$

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,

$E^o_{cell}=\frac{0.0592}{n}\times \log K$       ........(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

$-0.89v=\frac{0.0592}{6}\times \log K$

$\log K=-90.202$

$K=6.28\times 10^{-91}$

Therefore, the value of the equilibrium constant is, $K=6.27\times 10^{-91}$