A student measures the potential of a cell made up with I M CuSO_4 in one soluti

Question

A student measures the potential of a cell made up with I M CuSO_4 in one solution reservoir and I M ZnSO_4 in the other. There is a metallic copper (Cu^0) electrode in the CuSO_4 and a metallic electrode in the ZnSO_4, and the cell is set up as shown in Figure 32.1. She finds that the potential. voltage. of the cell, 1.076V. and that the Zn electrode is negative. At which electrode is oxidation occurring? Write the equation for the oxidation half-reaction in this cell. Write the equation for the reduction half-reaction in this cell. Write the net ionic equation for the spontaneous oxidation-reduction reaction that occurs in this cell. In another cell, the potential of the copper metal, copper(ll) ion electrode was found to be 0.442 V relative to a silver metal, silver ion electrode, with the copper electrode being negative. a. If the potential of the silver, silver ion electrode. is taken to be 0.000 V in oxidation or reduction, what is the value of the potential for the oxidation reaction. If equals + 0.799V, as in standard tables of electrode potentials, what is the value of the potential of the oxidation reaction of copper. The student adds 6 M NH, to the CuSO_4 solution in this second cell until the is essentially all converted. The voltage of the cell. electrode is still negative. Find the residual concentration of ion in the cell. (Use Eq. 4.)

Explanation / Answer

Answer – 1) We are given the cell with [CuSO4] = 1 M, [ZnSO4] = 1 M, Ecell = 1.076 V and Zn is electrode is negative

a)We know the oxidation means loss of electrons or increase in the oxidation state and we are given that Zn is electrode is negative. In the electrochemical cell the electrode with negative is anode and there is oxidation occur, so on the Zn-electrode oxidation occurring.

b) The oxidation half-reaction – We know at the Zn-electrode oxidation occurring, so the Zn metal gets oxidized as follow –

Zn(s) -----> Zn2+(aq) + 2e-

c) The reduction half-reaction – We know at the Zn-electrode oxidation occurring, so on Cu -electrode there is reduction occurring so reduction half-reaction as follow –

Cu2+(aq) + 2e- -----> Cu(s)

d) The net ionic equation for the spontaneous redox reaction as follow –

Zn(s) -----> Zn2+(aq) + 2e-

Cu2+(aq) + 2e- -----> Cu(s)

Zn(s) + Cu2+(aq) -----> Zn2+(aq) + Cu(s)

2) We are given, Ecell = 0.442 V, Cu is negative electrode

a)We know the negative electrode is anode and there is occurring oxidation, so on the Cu there is oxidation occurred. EoAg+ , Ag = 0.000 V in the reduction

so, EoCu , Cu2+ = ?

we know,

Eocell = Eooxd + Eored

First we need to calculate the Eocell for this cell

Cu(s) -----> Cu2+ + 2e-   , Eo = - 0.337 V

2Ag+ + 2 e- ----> 2 Ag(s) , Eo = +0.799 V

Cu(s) + 2Ag+ -----> Cu2+ + 2 Ag(s) , Eo = 0.462 V

Then Eocell = Eooxd + 0.00 V

So, EoCu , Cu2+ = Eocell

                     = 0.462 V

b) We are given, EoAg+ , Ag = 0.799 V

we know,

Eocell = Eooxd + Eored

0.462 V = EoCu , Cu2+ + 0.799 V

EoCu , Cu2+   = 0.462 -0.799

                     = -0.337 V

c) We are given, Ecell = 0.917 V , Eocell = 0.462 V

we know Nernst equation

Ecell = Eocell -0.0592/n ln K

0.917 V = 0.462 V – 0.0592 / 2 * ln [Cu2+] / [Ag+]

0.917 -0.462 = -0.0296 * ln [Cu2+] / 1.0

0.455 / -0.0296 = ln [Cu2+]

ln [Cu2+] = -15.4

taking anitln from both side

[Cu2+] = 2.10*10-7 M

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