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(i) Calculate DeltaG^(@) and log K_(c) for the following reaction at 298 K : 2Al(s)+3Cu^(2+)(aq)rarr 2Al^(3+)(aq)+3Cu(s) "Given :"E_("cell")^(@)=2.02V (ii) Using the E^(@) values of A and B, predict which is better for coating the surface of iron [E^(@)(Fe^(2+)//Fe)=-0.44V] to prevent corrosion and why? "Given : "E^(@)(A^(2+)//A)=-2.37V:E^(@)(B^(2+)//B)=-0.14V The conductivity of "0.001 mol L"^(-1) solution of CH_(3)COOH is 3.905 xx 10^(-5)" S cm"^(-1). Calculate its molar conductivity and degree of dissociation (alpha). "Given"lambda^(@)(H^(+))="349.6 S cm"^(2)"mol"^(-1) and lambda^(@)(CH_(3)COO^(-))="40.9 S cm"^(2)"mol"^(-1) What type of battery is dry cell? Write the overall reaction occurring in dry cell. |
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Answer» Solution :`DF^(@)=-nFE_("cell")^(@)` `DG^(@)=-6xx96500xx2.02` `E_("cell")^(@)=(0.059V)/(n)logKc` `logKc=(2.02Vxx6)/(0.059V)=205.42` A because is `E^(@)` value is more negative. `^^_(m)^(@)=KAPPA xx1000//C` `=3.905xx10^(-5)xx1000//0.001` `="39.05 S cm"^(2)//"mole"` `CH_(3)COOH rarr CH_(3)COO^(-)+H^(+)` `^^^(@)CH_(3)COOH=LAMBDA^(@)CH_(3)COO^(-)+lambda^(@)H^(+)` `=349.6+40.9` `^^^(@)CH_(3)COOH="390.5 S cm"^(2)//"mol"` `=0.1` Primary cell `Zn+2NH^(+)+2MnO_(2)rarr Zn^(++)+2NH_(3)+2MnO(OH)` Primary cell `Zn+2NH^(+)+2MnO_(2)rarr Zn^(++)+2NH_(3)+2MnO(OH)` |
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