Question

Rearrange the following half-reaction to create an electrochemical cell. Write the labelled cathode half-reaction, anode half-reaction, and net ionic reaction. Also calculate the net cell potential.

Cu(s )| Cu²⁺(aq) NO3^-(aq) || Na⁺(aq) SO4^2-(aq) | Pb(s)

Answer 1

CU(solid)/Cuso4(aq)//NaNo3(aq)/Na(solid),Pb this is rearrenged equation.

A. The anode half reaction.:where oxidation takes place hence the

metal will loose its electron And form cation

Hence anode half reaction will be Cu/Cuso4

Where Cu=2e-+Cu2+

Where cu solid metal will change in to aq cation form

B. The cathode half reaction At this electrode the metal will gain electron n always undergoes reduction process thereby reduction always place at anode

The reaction will be NaNo3(aq sol) =Na(solid)

C. The net ionic equation will be

cu/Cuso4//NaNo3/Na+No³⁺

^{d the net cell potentiol can be calculated by their standard elctrode potential which is not given there provide that information also.}

^{standard electrode of cu+}stanstandard electrode popotential of No3_{- ion}