1.

Calculate the potential for half cell containing 0.10 M K_(2)Cr_(2)O_(7)(aq),0.20" M "Cr^(3+)(aq) and 1.0xx10^(-4)M H^(+)(aq). The half-cell reaction is Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)+6e^(-)to2Cr^(3+)(aq)+7H_(2)O(l) and the standard electrode potential is given as E^(@)=1.33V.

Answer»


Solution :`E=E^(@)-(0.0591)/(n)"LOG"([Cr^(3+)]^(2))/([Cr_(2)O_(7)^(2-)][H^(+)]^(14))=1.33-(0.0591)/(6)"log"((0.20)^(2))/((0.10)(10^(-4))^(14))`
`=1.33-(0.0591)/(6)log(4xx10^(57))=1.33-(0.0591)/(6)(57+0.6021)=1.33-0.57=0.76V`


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