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Blue colour of copper sulphate is slowly discharged when an iron rod is dipped into it. Explain this by calculating DeltaG^(@)with the help of the following data : [E_(Cu^(2+)//Cu)^(@) = 0.34 V, E_(Fe^(2+)//Fe)^(@) = -0.44 V and 1 Faraday = 96500 C mol^(-1)] |
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Answer» Solution :`E_("cell")^(@) = E_("Cu^(2+)//Cu)^(@) - E_(Fe^(2+)//Fe)^(@) = +0.34 V - (-0.44 V) = +0.78 V` We have the equation `DeltaG^(@) =-nFE_("cell")^(@) = -2 xx 96500 C MOL^(-1) xx 0.78 V = -19301 J mol^(-1) = -19.3 kJ mol^(-1)` As the value of `DeltaG^(@)`is NEGATIVE, the reaction will take PLACE. That means BLUE colour of copper sulphate is slowly discharged. |
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