Label the diagram. Write the reduction half reaction, the oxidation half reaction and the net reaction as well as calculate the cell potential. ?

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Very confused..

1 Answer
Apr 5, 2018

See below

Explanation:

Looking at the notation you posted in the link:
I can say that
#Ba(s)# is oxidized to #Ba^(+2)(aq)#
#Pb ^(2+)(aq)# is reduced to #Pb(s)#

Anode is where oxidation occurs: #Ba(s)# electrode
Cathode is where reduction occurs: #Pb(s)# electrode

Electron flow, always from anode to cathode no matter if the cell is galvanic or electrolytic

Ion Movement:
1.Show that on the diagram by drawing an arrow from #Ba(s)# out to the #Ba^(+2)(aq)#
2. Show that on the diagram by drawing an arrow from #Pb^(+2)(aq)# out to the #Pb(s)#

Cell potential/ Half Reactions:
#color(white)(mmmmlmmmmmmmmmmmmmmmmmmmmll)ul(E^@"/V")#
#"Cathode": color(white)(m)"Pb"^(2+)"(aq)" + 2"e"^"-"color(white)## " Pb"^"""(s)" color(white)(mmmmmm)color(white)" -0.13#
#"Anode": color(white)(m)"Ba(s)" → "Ba"^"2+""(aq)" + 2"e"^"-"color(white)(mmmmmmmml) "+2.90"#
#"Cell":color(white)(mm)"Ba(s)" + "Pb"^"2+""(aq)" → "Ba"^"2+""(aq)" + "Pb(s)" color(white)(mm)"2.77"#

This is a spontaneous reaction in a galvanic cell