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fredd [130]
3 years ago
12

What is the value of the equilibrium constant for this redox reaction? 2Cr3+(aq) + 3Sn2+ (aq) 2Cr(s) + 3Sn4+(aq) E = -0.89 v

Chemistry
2 answers:
charle [14.2K]3 years ago
8 0
<span>b. square root of 3 over 3 is the answer to your question!!! I hope I helped!!!!!!!!!!                                                                 XoXo -Marcey<3!  :D
</span>
Kay [80]3 years ago
6 0

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}

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