Balance these equation

To balance the given equations, we need to ensure that the number of atoms for each element is the same on both sides of the reaction. Let's tackle these one at a time.

i) Al + Ag → Al²⁺ + Ag

This equation involves the oxidation of aluminum (Al) and the reduction of silver ions (Ag). The balanced half-reactions could look like this:

1. Oxidation half-reaction: \[ 2\text{Al} \rightarrow 2\text{Al}^{3+} + 6e^- \]

2. Reduction half-reaction: \[ 6\text{Ag}^+ + 6e^- \rightarrow 6\text{Ag} \]

Now, we can combine both half-reactions:

\[ 2\text{Al} + 6\text{Ag}^+ \rightarrow 2\text{Al}^{3+} + 6\text{Ag} \]

ii) MnO₂ + CrO₄²⁻ → Mn + CrO₄²⁻

For the second equation, we assume it is a redox reaction, involving the reduction of manganese dioxide (MnO₂) to manganese (Mn) and possibly the chromium species changing oxidation states as it involves dichromate (CrO₄²⁻):

1. Reduction half-reaction for manganese: \[ \text{MnO}_2 + 4\text{H}^+ + 2e^- \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O} \]

2. Oxidation half-reaction for chromium can be expressed if you want CrO₄²⁻ being converted to a more reduced form (like Cr³⁺), but for this scenario, it seems we are not provided with the full context to balance this completely without making assumptions on the products formed.

But to demonstrate a more simplified version of what we have, we'd need to confirm what the chromium product needs to be. If it's just staying Chrome in the form of \(\text{CrO}_4^{2-}\) then the reaction needs revision on what's happening exactly with chromium.

Given we've established manganese is reducing, add the electrons correctly collected from redox chemistry, currently, from above devoid of product shifting we'd observe:

1. Electrons gained/lost would be compared to establish an overall balance leading to recalibration through coefficients.

2. If you're dealing with acids/bases as a medium, we'd need to inflate the balancing further through said hydrogen ions or hydroxide ions accordingly.

If assuming CrO₄²⁻ doesn't undergo any shift: \[ \text{MnO}_2 + 4\text{H}^+ + 2e^- \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O} \]

2 \(\text{CrO}_4^{2-}\) sometime engages with 8H⁺ and 6e⁻ depending on final products since its balance leads back in equilibrium as observed in classical ionic representations.

Finally, ensure final counts are reconciled amongst both sides to abide by the law of conservation of mass, balancing electrons/states wherever needed.

Let me know if you want to delve deeper into the second reaction specifics if you're seeking particular oxidation states or outputs!