Question

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The question asks me to balance the following redox reaction using the half-reaction method: 10) Co(OH)3...

The question asks me to balance the following redox reaction using the half-reaction method:

10) Co(OH)₃ (s) + Sn (s) --> Co(OH)₂ (s) + HSnO₂^1- (aq) (basic solution)

Solutions

Expert Solution

Balance electron gain/loss

Co(OH)₃ + e^- --------> Co(OH)_2
- [ Co(III) to Co(II) ]

Balance O by adding H₂O

Co(OH)₃ + e^- --------> Co(OH)₂ + H₂O

Balance H by adding H⁺

Co(OH)₃ + e^- +H⁺ --------> Co(OH)₂ + H₂O -----(1)

Balance electron gain/loss

Sn ----> HSnO₂^1- + 2e^- [ Sn(0) to Sn(II)]

Balance O by adding H₂O

Sn + 2H₂O ----> HSnO₂^1- + 2e^-

Balance H by adding H⁺

Sn + 2H₂O ----> HSnO₂^1- + 2e^- + 3H⁺ ----(2)

Multiply equation (1) by 2

2Co(OH)₃ + 2e^- + ²H⁺ --------> 2Co(OH)₂ + 2H₂O -----(3)

Add equation (2) and (3)

Sn + 2H₂O ----> HSnO₂^1- + 2e^- + 3H⁺

2Co(OH)₃ + 2e^- +2H⁺ --------> 2Co(OH)₂ + 2H₂O

-------------------------------------------------------------------

2Co(OH)₃ + Sn --------> 2Co(OH)₂ + HSnO₂^1- + H⁺

-------------------------------------------------------------------

Since reaction occurs in basic solution add OH^- on both side to neutralize H⁺

2Co(OH)₃(s)+ Sn (s) + OH^- (aq) --------> 2Co(OH)₂(s) + HSnO₂^1- (aq)+ H₂O (l)

queen_honey_blossom answered 1 year ago

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