Balance each redox reaction using the half-reaction method.

Question

I have two questions, but they are the same type.

1. Zn(s)+Sn2+(aq)→Zn2+(aq)+Sn(s)
2. Mg(s)+Cr3+(aq)→Mg2+(aq)+Cr(s)

Answer 1

They probably want you to show the electron transfer as well as balancing the mass of each element.

1). Zn(s) + Sn²⁺(aq)==> Zn²⁺(aq) + Sn(s)

Sn²⁺ + 2e- ==> Sn(s) ... reduction half reaction

Zn(s) ===> Zn²⁺ + 2e- ... oxidation half reaction

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Zn(s) + Sn2+(aq) ==> Zn2+(aq) + Sn(s) ... balanced redox equation

2). Mg(s) + Cr³⁺(aq) ==> Mg²⁺(aq) + Cr(s)

Cr³⁺ + 3e- ==> Cr(s) ... reduction half reaction

Mg(s) ==> Mg²⁺ + 2e- ... oxidation half reaction

to equalize electrons, multiply reduction reaction by 2 and oxidation reaction by 3 to get ...

2Cr³⁺ + 6e- ==> 2Cr(s)

3Mg(s) ==> 3Mg²⁺ + 6e-

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2Cr³⁺(aq) + 3Mg(s) ==> 2Cr(s) + 3Mg²⁺(aq) ... balanced redox

Answer 2

You can easily reach out to [email protected] if you have any more questions.

Let's look at the second one being the most challenging.

Step 1. Identify oxidation and reduction reaction

Cr³⁺(aq) + 3e- → Cr_(s) ; Chromium gains 3e- and is therefore reduced

Mg_(s) → Mg²⁺_(aq) + 2e- ; Magnesium loses 2e- and is therefore oxidized

Step 2. Balance electrons

By balancing electrons you can easily balance the overall equation.

2*(Cr³⁺(aq) + 3e- → Cr_(s)) 2Cr³⁺(aq) + 6e- → 2Cr_(s)

3*(Mg_(s) → Mg²⁺_(aq) + 2e- ) 3Mg_(s) → 3Mg²⁺_(aq) + 6e-

Step 3. Add the half reactions

3Mg_(s) + 2Cr³⁺_(aq) → 2Cr_(s) + 3Mg²⁺_(aq)

Answer 3

The super fast way to approach this type of balancing is to make sure the charge is the same on both sides:

The first question already has the charge the same so it is balanced!

1. Zn(s)+Sn2+(aq)→Zn2+(aq)+Sn(s)

THe second one below does not have charge balanced - the left is +3 and the right is +2 so you would have to multiply the Cr3+ by 2 and Cr by 2 and the Mg2+ by 3 and Mg by 3 this would make the charge on both sides +6 overall and balanced

1. 3Mg(s)+2Cr3+(aq)→3Mg2+(aq)+2Cr(s)