Show all work.Balance each of the following half-reactions under basic conditions (#1)Cr^3+(aq)-->CrO4^2-(aq) (#2)HSnO2^-(aq)-->HSnO3^-(aq)?

Cr{3+} → Cr{6+} + 3 e{-}
Cr{3+} → CrO4{2-} + 3 e{-}
Cr{3+} + 8 OH{-} → CrO4{2-} + 3 e{-}
Cr{3+} + 8 OH{-} → CrO4{2-} + 4 H2O + 3 e{-}

Sn{4+} → Sn{6+} + 2 e{-}
HSnO2{-} → HSnO3{-} + 2 e{-}
HSnO2{-} + 2 OH{-} → HSnO3{-} + 2 e{-}
HSnO2{-} + 2 OH{-} → HSnO3{-} + H2O + 2 e{-}

(#1)
Cr³⁺(aq) → CrO₄²⁻(aq)

1. Cr is balanced as there is 1 Cr on each side.

2. There are 4 more O on the right. Add 8OH⁻ to the left and 4H₂O to the right so that there are 8 H and 8 O on each side.
Cr³⁺(aq) + 8OH⁻(aq) → CrO₄²⁻(aq) + 4H₂O(l)

3. There are totally 5 negative charges on the left and 2 negative charges on the right (OR: The oxidation number of Cr increases by 3). Add 3 e⁻ to the right so that there are 5 negative charges on each side. The balanced half-equation is :
Cr³⁺(aq) + 8OH⁻(aq) → CrO₄²⁻(aq) + 4H₂O(l) + 3e⁻


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(#2)
HSnO₂⁻(aq) → HSnO₃⁻(aq)

1. Sn is balanced as there is 1 Sn on each side.

2. There are 1 more O on the right. Add 2OH⁻ to the left and H₂O to the right so that there are 3 H and 4 O on each side.
HSnO₂⁻(aq) + 2OH⁻(aq) → HSnO₃⁻(aq) + H₂O(l)

3. There are totally 3 negative charges on the left and 1 negative charge on the right (OR: The oxidation number of Sn increases by 2). Add 2 e⁻ to the right so that there are 3 negative charges on each side. The balanced half-equation is :
HSnO₂⁻(aq) + 2OH⁻(aq) → HSnO₃⁻(aq) + H₂O(l) + 2e⁻