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Answer 1 · 2015-05-15T23:10:54

$E^(@)=+0.37"V"$

$DeltaG^(@)=-35.7"kJ"$

$K_c=1.82xx10^(6)"mol"^(-1)."l"$

Start with the $E^(@)$ values, listing them in increasing order:

$Cu_((aq))^(2+)+erightleftharpoonsCu_((aq))^(+)$ $E^(@)=+0.153"V"$ $color(red)((1))$

$Cu_((aq))^(+)+erightleftharpoonsCu_((s))$ $E^(@)=+0.52"V"$ $color(red)((2))$

To work out $E_((cell))^(@)$ subtract the least +ve value from the most +ve.:

$E_((cell))^(@)=+0.52-0.153=+0.37"V"$

This means $color(red)((2))$ will be driven left to right and $color(red)((1))$ will be driven right to left $rArr$

$2Cu_((aq))^(+)rarrCu_((aq))^(2+)+Cu_((s))$

$DeltaG^(@)=-nFE_((cell))^(@)$

$= -1xx9.65xx10^(4)xx0.37=-35705"J"$

$=-35.7"kJ"$

$DeltaG^(@)=-RTlnK_c$

$lnK_c=(-DeltaG)/(RT)$

$lnK_c=(35.7xx10^(3))/(8.31xx298)=14.416$

From which:

$K_c=1.82xx10^(6)"mol"^(-1)."l"$

This is an example of a disproportionation reaction where a species, in this case copper(1), is simultaneously oxidised and reduced.