InterviewSolution
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InterviewSolution
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A Voltaic cell is set up at `25^(@)C` with the following half cells ? |
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Answer» `K = (1)/(R) xx ((1)/(A))` `^^_(m) = (K xx 1000)/(M)` (`{:("at anode": , Al(s) to Al^(3+) (aq) + 3e^(-)"]" xx 2), ("at cathode" , Ni^(2+)(aq) + 2e^(-) to Ni(s) xx 3):}/(2Al(s) + 3Ni^(2+)(aq) to Al^(3+) (aq) + Ni(s)))/` `E_("cell") = E_("cell")^(0) - (0.0591)/(n) "log" ([Al^(3+)]^(2))/([Ni^(2+)]^(3))` `n = 6 , [Al]^(3+) = 0.001 M = 1 xx 10^(-5) M` `[Ni^(2+)] = 0.5 M E^(@) Ni^(2+)//Ni - E^(@) Al^(3+)//Al = -0.25 V - (-1.66V)` `E_("cell")^(@) = 1.41 V` `E_("cell")^(@) = 1.41 - (0.0591)/(6) "log" ((10^(-3))^(2))/((0.5)^(3)) = 1.41 - (0.0591)/(6)"log" (10^(-6))/(0.125)` `= 1.41 - (0.0591)/(6) "log" (10^(-6) xx 8) = 1.41 - (0.0591)/(6) ("log" 10^(-6) + "log" 2^(3))` `= 1.41 - (0.0591)/(6) (-6 "log" + 3 "log" 2)` `= 1.41 - (0.0591)/(6) (-6 + 3 xx 0.3010) = 1.41 - (0.0591)/(6) (-5.097)` `= 1.41 + (0.3012)/(6) = 1.46V` |
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