Consider the following electrochemical cell:
Al(s) | Al^3+(aq)(0.155 M) || H^(aq)
(0.00233 M), MnO4^-(aq)(0.0377 M), Mn^2+(aq)
(0.0168 M) | C (graphite)
Write a balanced equation for the redox reaction taking place in this cell. i am having trouble with the cathode half reaction.
MnO4^- + 8H^+ + 5e ==> 4H2O + Mn^2+
To write the balanced equation for the redox reaction occurring in the given electrochemical cell, we need to determine the cathode half-reaction.
In this cell setup, the anode half-cell is Al(s) | Al^3+(aq) (0.155 M), and the cathode half-cell contains H^+(aq) (0.00233 M), MnO4^-(aq) (0.0377 M), and Mn^2+(aq) (0.0168 M) | C (graphite).
To identify the cathode half-reaction, we can look for species being reduced or gaining electrons. In this case, we see that MnO4^- is being reduced to Mn^2+. Let's write the half-reaction for the reduction of MnO4^-:
MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O
Now, we need to balance the equation. We have eight hydrogen ions (H^+) on the left side and only four on the right side. To balance the hydrogen atoms, we can add four water molecules (H2O) to the left side:
MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O
Next, we balance the oxygen atoms. On the left side, we have four oxygen atoms from MnO4^-, and on the right side, we have four oxygen atoms from water molecules. Therefore, the equation is already balanced.
Finally, to make sure the number of electrons is equal on both sides, we verify that we have five electrons (5e^-) on the left side.
Therefore, the balanced cathode half-reaction is:
MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O