Question

In: Chemistry

Cd (s) + NiO(OH) (s) -> Cd(OH)2 (s) + Ni(OH)2 (not balanced) a. Balance the reaction...

Cd (s) + NiO(OH) (s) -> Cd(OH)2 (s) + Ni(OH)2 (not balanced)

a. Balance the reaction above usingt the 1/2 reaction method, in alkaline conditions

b.`if 5.00 g of Cd and 7.5 g of NiO(OH) are combined, which will be the limiting reactant? How much of the other reactant will remain after the reaction completes?

Solutions

Expert Solution

Cd (s) + NiO(OH) (s) -> Cd(OH)2 (s) + Ni(OH)2

Step 1: Separate the two half-reactions.
Oxidation : Cd (s) -----> Cd(OH)2
Reduction : NiO(OH) (S)----> Ni(OH)2

Step 2: First balance the elements other than O and H.
Ni and Cd both are already balanced.

Step 3: Balance the oxygens using H2O
Oxidation : Cd (s) + 2 H2O -----> Cd(OH)2
Reduction : NiO(OH) (S)----> Ni(OH)2

Step 4: Balance the hydrogens using protons (H+)

Oxidation : Cd (s) + 2 H2O -----> Cd(OH)2 + 2 H+
Reduction : NiO(OH)(s) + H+ ----> Ni(OH)2

Step 5: Now, balance the charge with electrons.
Oxidation : Cd (s) + 2 H2O -----> Cd(OH)2 + 2 H+ + 2e-
Reduction : NiO(OH)(s) + H+ + e- -----> Ni(OH)2

Step 6: Scale the reactions so that the electrons are equal.
Oxidation : Cd (s) + 2 H2O -----> Cd(OH)2 + 2 H+ + 2e-
Reduction : 2NiO(OH)(s) + 2H+ + 2e- -----> 2Ni(OH)2

Since, the reaction is in alkali condition, to remove H+ ions we need to add OH- ions.
Therefore, the reactions will become;
Oxidation : Cd (s) + 2 H2O + 2OH- -----> Cd(OH)2 + 2 H+ + 2OH- + 2e-
Reduction : 2NiO(OH)(s) + 2H+ + 2OH- + 2e- -----> 2Ni(OH)2

Since, H+ OH- = H2O; Therefore;
Oxidation : Cd (s) + 2OH- -----> Cd(OH)2 + 2e-
Reduction : 2NiO(OH)(s) + 2H2O + 2e- -----> 2Ni(OH)2

Step 7: Add the reactions and cancel out common terms.
Cd(s) + 2H2O + 2NIO(OH) (s) -----> Cd(OH)2 + Ni(OH)2

Hence, the balanced reactions is: Cd(s) + 2H2O + 2NIO(OH) (s) -----> Cd(OH)2 + Ni(OH)2

b) Balance reaction: Cd(s) + 2H2O + 2NIO(OH) (s) -----> Cd(OH)2 + Ni(OH)2

1 mole of Cd reacts with 2 mol of NiO(OH) and produced 1 mol of Cd(OH)2 and 1 mol of Ni(OH)2.

no. moles of Cd = mass/mol.wt
               = 5.00 g/ 112.411 g/mol
               = 0.0445 mol
No. of moles of NiO(OH) = mass/mol.wt      
                       = 7.5 g/91.7001 g/mol
                       = 0.0818 mol
                      
                      
Since 0.0818 mol of NiO(OH) require 0.0409 mol of Cd. So, we have an excess of Cd present.
Therefore, the limiting reagent is NiO(OH).

How much of the other reactant will remain after the reaction completes?

Remaining moles of Cd = 0.0445 mol - 0.0409 mol
                      = 0.0036 mol
Mass of Cd reaming unreacted = no. moles x mol.wt
                           = 0.0036 mol x 112.411 g/mol
                           = 0.405 g
Hence, the mass of Cd reamining after completion of reaction = 0.405 g


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