Question

Write a balanced equation from each line notation:

a) Ag(s)|Ag¹ᐩ(aq)||Cr³ᐩ(aq)|Cr(s)

b)Pb(s)|Pb²(aq)||MnO₂(aq)|Mn²ᐩ(aq)|Pts)

Answer 1

Equation (a)

First half reaction

Ag(s) --------------> Ag⁺(aq) +e ^-

Second half reaction

Cr³⁺(aa) + 3e^- ------->Cr(s)

Multiplying 3 from first half reaction than adding second half reaction (electron is equal and opposite side in two reaction that's why electron is cencle)

3Ag(s) + Cr³⁺(aq) --------> 3Ag⁺(aq) + Cr(s)

Part(b)

Step (A)
Pb(s) → Pb²⁺(aq) (oxidation)

MnO₂→ Mn²⁺(aq) (reduction)

Step B
Balance each half reaction.

Pb Half Reaction: To balance this equation, we simply need to add two electron to the right side of the equation.

Pb → Pb²⁺(aq) + 2e-

Mn Half Reaction: We add water to the right to balance the two oxygens on the left.

MnO2(aq) → Mn2+(aq) + 2H2O

Next, to balance the 4 hydrogens in the water, we add 4H+(aq) on the left.

MnO2(aq) + 4H+(aq) → Mn2+(aq) + 2H2O

Now, looking at the equation above, we can see the charges are unbalanced on the left vs the right side. The overall charge on the left is +4 and on the right is +2. To balance the charge, we need to add 2 electrons to the left side of the equation.

MnO(aq) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O

Step C.

Add the half reactions. The electrons on each side cancel to give the overall reaction:

Pb(s) + MnO2(aq) + 4H+(aq) → Pb2+(aq) + Mn2+(aq) + 2H2O