InterviewSolution
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InterviewSolution
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Consider the following four electrodes: `P=Cu^(2+)(0.0001M)//Cu(s)" "Q=Cu^(2+)(0.1M)//Cu(s)` `R=Cu^(2+)(0.01M)//Cu(s)" "S=Cu^(2+)(0.001M)//Cu(s)` If the standard electrode potential of `Cu^(2+)//Cu` is +0.34V, the reduction potentials in volts of the above electrodes follow the order:A. PgtSgtRgtQB. SgtRgtQgtPC. RgtSgtQgtPD. QgtRgtSgtP |
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Answer» Correct Answer - D `Cu^(2+)+2e^(-)toCu,E_(red)=E_(red)^(@)+(0.0591)/(2)log[Cu^(2+)]` Thus, greater is `Cu^(2+)` ion concentration, greater will be the reduction potential. |
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