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Calculate the voltage E of the cell at 2...

Calculate the voltage `E` of the cell at `25^(@)C`
`Mn(s) |Mn(OH_(2))(s)|Mn^(2+)(x,M)OH^(-) (1.00 xx 10^(-4)M)||Cu^(2+) (0.675M)|Cu(s)`
given that `K_(sp) = 1.9 xx 1-^(-13)` for `Mn(OH)_(2)(s) E^(@) (Mn^(2+)//Mn) =- 1.18 V, E^(@) (Cu^(+2)//C) = +0.34V`

Text Solution

Verified by Experts

The correct Answer is:
`1.66V`
`Mn rarr Mn^(2+) +2e^(-)`
`Cu^(2+) +2e^(-) rarr Cu K_(sp) = [Mn^(2+)] [OH^(-)]^(2)`
`E = (1.18 +0.34) - (0.0591)/(2) log. ([Mn^(2+)])/([Cu^(2+)])`
`E_(cell) = 1.66V`
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