Correction:
E_cell= -0.23 - 2.37= -2.60 V
Calculate the standard cell potential for each of the following electrochemical cells.
Ni^+2(aq)+Mg(s)--->Ni(s)+Mg^+2(aq)
I used the e cell values from the back of my book. They are:
Ni^+2(aq)+2e--->Ni(s) E^0= -0.23 V
Mg^+2(aq)+2e--->Mg(s) E^0= -2.37
I did E^0= -0.23 - 2.37= -2.60 V ; however, my online hw said that was wrong. Is there another step I'm missing?
3 answers
Look at your complete equation.
Ni^+2(aq)+Mg(s)--->Ni(s)+Mg^+2(aq)
Ni^+2 + 2e ==> -0.023 v
Mg(s) ==> Mg^+2 + 2e is the reverse of the cell you looked up; therefore, the voltage for this cell is +2.37. Then Ecell = -0.23 + 2.37 = ??
(You have added a reduction half cell to an oxidation half cell.)
Ni^+2(aq)+2e--->Ni(s) E^0= -0.23 V
Mg^+2(aq)+2e--->Mg(s) E^0= -2.37
Ni^+2(aq)+Mg(s)--->Ni(s)+Mg^+2(aq)
Ni^+2 + 2e ==> -0.023 v
Mg(s) ==> Mg^+2 + 2e is the reverse of the cell you looked up; therefore, the voltage for this cell is +2.37. Then Ecell = -0.23 + 2.37 = ??
(You have added a reduction half cell to an oxidation half cell.)
Ni^+2(aq)+2e--->Ni(s) E^0= -0.23 V
Mg^+2(aq)+2e--->Mg(s) E^0= -2.37
2.14 V
4 answers