Write the half reactions and the balanced
equation for the galvanic cell
Hg(ℓ)| Hg2+
2
(aq)||
MnO−
4
(aq), Mn2+(aq), H
+(aq)| Pt(s).
What is the smallest possible integer coef-
ficient of H+(aq) in the combined balanced
equation?
Also...
Hg2(2+) + 2e- >> Hg2(2+) +0.92V
Mn(2+) + 2e- >> Mn -1.18V
Don't cheat!
To determine the half reactions and the balanced equation for the given galvanic cell, we need to understand the oxidation and reduction reactions occurring at each electrode.
The anode (oxidation half-reaction):
In this case, the anode is represented by the Hg(ℓ)| Hg2+(aq) electrode. Here, the Hg(ℓ) is being oxidized to Hg2+(aq) while losing two electrons. The half-reaction can be written as:
Hg(ℓ) ⟶ Hg2+(aq) + 2e-
The cathode (reduction half-reaction):
The cathode involves the MnO−4(aq), Mn2+(aq), H+(aq)| Pt(s) electrode. Here, MnO−4(aq) is being reduced to Mn2+(aq) by gaining five electrons. The half-reaction can be written as:
MnO−4(aq) + 8H+(aq) + 5e- ⟶ Mn2+(aq) + 4H2O(l)
To balance the half-reactions, we need to ensure that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction. Multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2 will balance the electrons:
5Hg(ℓ) ⟶ 5Hg2+(aq) + 10e-
2MnO−4(aq) + 16H+(aq) + 10e- ⟶ 2Mn2+(aq) + 8H2O(l)
Now, we can combine these half-reactions to write the overall balanced equation. To do this, we need to ensure that the number of electrons gained and lost cancel out. To cancel out the electrons, we can multiply the first half-reaction by 2 and the second half-reaction by 5:
10Hg(ℓ) + 2MnO−4(aq) + 16H+(aq) ⟶
10Hg2+(aq) + 4Mn2+(aq) + 8H2O(l)
The smallest possible integer coefficient of H+ in the combined balanced equation is 16.
Therefore, the balanced equation for the galvanic cell is:
10Hg(ℓ) + 2MnO−4(aq) + 16H+(aq) ⟶ 10Hg2+(aq) + 4Mn2+(aq) + 8H2O(l)