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The `E^(0)(M^(2+)//M)` value for copper is positive `(+0.34V)`. What is possibly the reason for this? |
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Answer» `E^(ɵ)((M^(2+))/(M))` for any metal is related to the sum of the enthalpy changes taking place in the following steps: `M_(s)overset(triangle_aH^(ɵ))toM_(g)(triangle_aH^(ɵ)=` enthalpy of atomisation) `M_(g)overset(triangle_iH^(ɵ))toM_(g)^(2+)(triangle_iH^(ɵ)=`ionisation enthalpy) `M_(g)^(2+)+aq.overset(triangle_(hyd)H^(ɵ))toM_(aq)^(2+)(triangle_(hyd)H^(ɵ)=`hydration enthalpy) Copper has high (positive) enthalpy of atomisation and low (negative) enthalpy of hydration. Hence `E^(ɵ)((Cu^(2+))/(Cu))` is positive. |
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