Cel,l Metal at the (+) cathode ,metal at the (-) anode, cell potential (volts) a

Question

Cel,l Metal at the (+) cathode ,metal at the (-) anode, cell potential (volts)

a. Mg-Cu, Cu, Mg, 1.672 V

b. Cu-Zn, Cu, Zn, 1.100V

C. Mg-Zn, Zn, Mg, 0.517V

*******************Please show work I want to see how you come up with the correct answer**********************

1. use the half ceell potention for copper, the voltage observed for the Mg-Cu electrochemical cell, and Eq

Ecel = E red (cathode) - E red (anode) to calculate the half-cell reduction potential for an Mg/Mg 2+

2. Use the calculated Mg/Mg 2+ half cell potential and the coltage observed for the Mg-Zn electochemical cell to calculate the half-cell reduction potential of the Zn/Zn 2+ half-cell

3. Write the thre half-cell reactions as reductions with their corresponding reduction potential voltages, listing hte metal that is most easily oxidized first

first:

Second:

third:

Explanation / Answer

1. E0Cell = E0cathode - E0anode

cu =+ 0.34 v

1.672 = 0.34- Mg

Mg =0.34 - 1.672 = -1.332 v

2. Mg=-1.332 v

Ecell= 0.517 v

Ecell= Ecathode - Eanode

0.517 = -1.332-(Zn)

Zn = -1.332-0.517 = -1.849 v

3. i. Cu+2(aq) + 2 e- ---> Cu(s)    Ered= + 0.34 v

ii. Mg+2(aq) + 2 e- ---> Mg(s)    Ered= -1.332 v

iii. Zn+2(aq) + 2 e- ---> Zn(s)    Ered= -1.889 v

Easily oxidise metal having higher reduction potential. so that order of metal towards oxidation is

Cu(First)> Mg(2nd)> Zn(3rd)

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