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The electrode potentials for `Cu^(2+)+e^(-)toCu^(+)` and `Cu^(+)e^(-)toCu_(s)` are `+0.15`V and `+0.50V` respectively the value of `E_((Cu^(2+))/(Cu))^(@)` will be?A. 0.500 VB. 0.325 VC. 0.650 VD. 0.150 V |
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Answer» Correct Answer - B `Cu^(2+)+1e^(-)toCu^(+)` `E_(1)^(0)=0.15vDeltaG_(1)^(0)=-n_(7)E_(1)^(0)F` underline(Cu^(+)+e^(-)toCu_(s) E_(2)^(0)=0.50vDeltaG_(2)^(0)=-n_(7)E_(2)^(0)F)` `Cu^(2+)+2e^(-)toCuDeltaG^(@)=DeltaG_(1)^(@)+DeltaG_(2)^(@)` `(-1)nEF=(-1)n_(1)E_(1)^(0)F+(-1)n_(2)E_(2)^(0)F` `E=(n_(1)E_(1)^(0)+n_(2)E_(2)^(0))/(n)=(0.15xx1+0.50xx1)/(2)=0.325V` |
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