Question

Suppose you made a copper-nickel cell where the copper was the cathode and the nickel was the anode. You add a 0.5000 M solution of copper nitrate on the cathode side and a 3.000 M solution of nickel nitrate on the anode side. What is the voltage of the cell under these conditions? The units are volts and use at least 4 significant digits in the answer. The error interval is 0.02 V.

Answer 1

anode reaction: oxidation takes place

Ni(s) -------------------------> Ni+2 (aq) + 2e-   ,   E⁰_Ni+2/Ni = - 0.26 V

cathode reaction : reduction takes palce

Cu+2 (aq) + 2e- -----------------------------> Cu(s) , E⁰_Cu+2/Cu = 0.34 V

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net reaction: Ni(s) +Cu+2(aq) -------------------------> Ni+2 (aq) + Cu(s)

E⁰_cell= E⁰_cathode- E⁰_anode

E⁰_cell= E⁰_Cu+2/Cu - E⁰_Ni+2/Ni

          = 0.34 - (-0.26)

          = 0.6000 V

nernest equation

Ecell = E⁰_cell -2.303RT/nF* log [Ni+2]/[Cu+2]

Here R= universal gas constant 8.314 J/K mol

T = absolute temperature =25(0C)= 298k

F= faraday = 96500 Coloumb/mol

n   = no of moles of electrons are transfered =2

2.303RT/F= 0.0591

Ecell = E⁰_cell -(0.0591/n)* log [Ni+2]/[Cu+2]

Ecell = 0.6000 - (0.059 x1/2) *log 3.0/0.5}

         = 0.5770

Ecell = 0.5770