A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 deg

Question

A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 degrees C. The initial concentrations of Ni2+ and Zn2+ are 1.60 M and 0.130 M, respectively. I found the initial cell potential: 0.56 V What is the concentration of Ni2+ when the cell potential falls to 0.46 V? What is the concentration of Sn2+ when the cell potential falls to 0.46 V? I know you have to use the Nernst equation and solve for concentration, but I don't understand how I can find the concentration of one without knowing the concentration of the other. Presumably both change when the voltage changes. Thanks for your help!!

Explanation / Answer

E ( Complete Cell ) = E ( reduction of Cathode ) - E ( reduction of Anode ) E ( reduction of Cathode ) = Eo ( Cu^2+ : Cu ) - ( 0.059 / 2 ) log ( 1 / [ Cu^2+ ] ) = ( + 0.34 ) - ( 0.059 / 2 ) log ( 1 / [ 0.250 ] ) E ( reduction of Anode ) = Eo ( Pb^2+ : Pb ) - ( 0.059 / 2 ) log ( 1 / [ Pb^2+ ] = ( - 0.13 ) - ( 0.059 / 2 ) log ( 1 / [ 5.30 x 10^-2 ] ) E ( Complete Cell ) = ( + 0.34 ) - ( - 0.13 ) - { ( 0.059 / 2 ) log ( [ 5.30 x 10^-2 ] / [ 0.250 ] ) } = ( + 0.47 ) - { ( 0.059 / 2 ) log ( [ 5.30 x 10^-2 ] / [ 2.50 x 10^-1 ] ) } = ( + 0.47 ) - ( 0.059 / 2 ) { ( log 5.30 - 2 ) - ( log 2.50 - 1 ) } = ( + 0.47 ) - ( 0.059 / 2 ) { ( 0.7243 ) - ( 0.3797 ) -1 } = ( + 0.47 ) - { ( 0.059 / 2 ) x ( - 0.6554 ) } = ( + 0.47 ) - ( - 0.0193343 ) = + 0.4893343 ( Rounded up to + 0.49 V ) It is possible that the answer . . . . may need the + or - sign in front of the value may not need the unit V behind the value For your other question , ( Electrolysis of CuBr2 ) your electron balancing concept is correct . You should try different format - - or - - Check whether to represent reactant as CuBr2(aq) [[ this means that CuBr2 does not ionize and the reactants not exist as Cu++(aq) / Cu2+(aq) / Cu^2+(aq) or Br-(aq) / Br^-(aq) ]] CuBr2(aq) + 2e- -> Cu(s) + 2Br-(aq) __ Cathode CuBr2(aq) -> Br2(aq) + 2e- + Cu2+(aq) __ Anode or CuBr2(aq) -> Br2(l) + 2e- + Cu2+(aq) __ Anode Note : It is already " 1 Year " that I joined this Yahoo Answers . During this 1 year , I never answer questions related to Nernst Equation . This is my first answer for this topic . You should also notice that my calculation method is a little bit different from your method . This calculation method will be more useful if you also want to calculate the equilibrium constant or concentration of ions / molecules at equilibrium .

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