A voltaic cell that uses reaction: PdCl4^2-(aq)+Cd(s)-->Pd(s)+4Cl-(aq)+Cd^2+(aq)

Question

A voltaic cell that uses reaction: PdCl4^2-(aq)+Cd(s)-->Pd(s)+4Cl-(aq)+Cd^2+(aq) has a measured standard cell potential of +1.03V 1. Write the two half-cell reactions. 2. determine E red for the the reaction involving Pd 3. Sketch the voltaic cell, label anode and cathode, and indicate the direction of electron flow

Explanation / Answer

Ni2+(aq) + 2 e --> Ni(s) Eo = –0.257 Zn(s) --> Zn2+(aq) + 2 e Eo = +0.763 so Zn(s) + Ni2+(aq) ? Zn2+(aq) + Ni(s) has Eo = 0.506 volts & n = 2 moles of electrons ===================================== nernst equation E = Eo - (0.0592 /n) (log Q) E = Eo - (0.0592 /n) (log Q) E = 0.506 - (0.0592 / 2) (log [products] / reactants] E = 0.506 - (0.0296) (log [Zn+2] / [Ni+2] a) What is the emf of this cell when [Ni2+ ] = 3.49 M and [Zn2+ ] = 0.106 M? E = 0.506 - (0.0296) (log [Zn+2] / [Ni+2] E = 0.506 - (0.0296) (log [0.106] / [3.49] E = 0.506 - (0.0296) (log 0.0304) E = 0.506 - (0.0296) (-1.52) E = 0.506+ 0.0449 your first answer is E = 0.551 volts ======================================… (b) What is the emf of the cell when [Ni2+ ] = 0.122 M and [Zn2+ ] = 0.923 M? E = 0.506 - (0.0296) (log [Zn+2] / [Ni+2] E = 0.506 - (0.0296) (log [0.923] / [0.122] E = 0.506 - (0.0296) (log 7.566) E = 0.506 - (0.0296) (0.879) E = 0.506 - 0.026 your first answer is E = 0.480 volts be sure to compare my original 1/2 reaction cell voltages with yours books rarely agree, & yours may have had a different # of sig figs

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