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?
2.14 V
To calculate the standard cell potential (E°) for an electrochemical cell, you need to consider the following two steps:
1. Identify the half-reactions: Divide the given chemical equation into two half-reactions, one for the reduction and one for the oxidation. In this case, the reduction half-reaction is Ni^+2(aq) + 2e^- → Ni(s) and the oxidation half-reaction is Mg(s) → Mg^+2(aq) + 2e^-.
2. Combine the half-reactions: Before combining the half-reactions, you need to ensure that the number of electrons transferred in each half-reaction is equal. In this case, both half-reactions involve the transfer of 2 electrons, so no balancing is required. Now, reverse one of the half-reactions so that the electrons cancel out when summed together. Reversing the oxidation half-reaction gives Mg^+2(aq) + 2e^- → Mg(s).
Now, you can add the two half-reactions together:
Ni^+2(aq) + 2e^- → Ni(s)
Mg^+2(aq) + 2e^- → Mg(s)
-------------------------------
Ni^+2(aq) + Mg(s) → Ni(s) + Mg^+2(aq)
Once you have the overall balanced equation, you can add the standard reduction potentials (E°) for the two half-reactions to determine the standard cell potential (E°cell). However, since the reduction potentials are given already, you just need to sum them up without any further calculations:
E°cell = E°reduction (cathode) - E°oxidation (anode)
= (-0.23 V) - (-2.37 V)
= 2.14 V
So, the correct standard cell potential for the given electrochemical cell is 2.14 V.
Therefore, you made a mistake in your calculation. Instead of subtracting the E° values, you need to subtract the E° of the oxidation from the E° of the reduction.
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
Correction:
E_cell= -0.23 - 2.37= -2.60 V