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The change in potential of the half-cell...

The change in potential of the half-cell Cu 2+ /Cu, when aqueous Cu2+ solution is diluted 100 times at 298 K? `(2. 303 RT)/(F) = 0.06

A

increases by 120 mV

B

decreases by 120 mV

C

decreases by 60 mV

D

no change

Text Solution

Verified by Experts

The correct Answer is:
D
Half cell reaction is
`Cu ^(2+) + 2 e ^(-) to Cu , n =2`
From Nernst equation.
`E = E ^(@) - ( 2.303 RT)/(nF) log "" (1)/([ Cu ^(2+)])`
`E _(1) = E ^(@) + (0.06)/(2) log [ Cu ^(2+) ]`
Let the initlal concentration of `Cu ^(2+) be 1.`
`E_(1) = E ^(@) + (0.06)/(2) log 1= E ^(@) + 0 therefore E _(1 ) = E ^(@)`
Further the `[Cu ^(2+)]` solution is dilued to 100 times.
`therefore underset("Initial") (M _(1) V _(1)) = underset("After dilution") ( M _(2) V _(2))`
`1 xx1 = M _(2) xx 100`
`M _(2) = (1)/(100) = 0.01`
`therefore E _(2) = E ^(@) + (0.059)/(2) log [0.01]`
`= E ^(@) + ( 0.059)/(2) (-2)`
`= E _(1) - 0.059 V = E _(1) - 59 mV`
Thus, the potential decreases by `59 (~~60) mV.`
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