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Calculate the equilibrium constant for t...

Calculate the equilibrium constant for the reaction:
`3Sn(s) +2Cr_(2)O_(7)^(2-) +28 H^(+) rarr 3Sn^(+4) +4Cr^(3+) +14H_(2)O`
`E^(@)` for `Sn//Sn^(+2) = 0.136 V E^(@)` for4 `Sn^(2+)//Sn^(4+) =- 0.154V`
`E^(@)` for `Cr_(2)O_(7)^(2-)//Cr^(3+) = 1.33V`

A

`10^(413)`

B

`10^(1.8)`

C

`10^(272)`

D

`10^(317)`

Text Solution

Verified by Experts

The correct Answer is:
C
Number of electrons exchanged `= 12` for 3Sn)
`E_(cell)^(@) = 0.009 + 1.33 = 1.339 = (0.0591)/(12)"log" K_(eq).`
`K_(eq) = 10^(271.878)`
`= 10^(272)`
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