Question
In: Chemistry
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) + HSnO21- (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 ----> HSnO21- + 2e- [ Sn(0) to Sn(II)]
Balance O by adding H2O
Sn + 2H2O ----> HSnO21- + 2e-
Balance H by adding H+
Sn + 2H2O ----> HSnO21- + 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 ----> HSnO21- + 2e- + 3H+
2Co(OH)3 + 2e- +2H+ --------> 2Co(OH)2 + 2H2O
-------------------------------------------------------------------
2Co(OH)3 + Sn --------> 2Co(OH)2 + HSnO21- + 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) + HSnO21- (aq)+ H2O (l)
queen_honey_blossom answered 2 years agoRelated Solutions
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