Answer:
- <em>Oxidation half-reaction</em>:
Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻
- <em>Reduction half-reaction</em>:
Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)
Explanation:
The reaction that takes place is:
- Fe²⁺(aq) + Ce⁴⁺(aq) → Fe³⁺(aq) + Ce³⁺(aq)
The <em>oxidation half-reaction</em> is:
- Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻
It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.
The <em>reduction half-reaction</em> is:
- Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)
It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.
<span>the theoretical yield which is the expected yield and the actual yield obtained are not always the same. therefore percent yield is calculated which shows how much of the percentage of the theoretical yield is actually obtained.
the theoretical yield = 56.0 g
actual yield = 47.0 g
percent yield = actual yield / theoretical yield x 100 %
percent yield = 47.0 / 56.0 x 100% = 83.9 %
percent yield = 83.9 %</span>
The value of equilibrium constant is equal to the quotient of the products raised to its stoichiometric coefficient over the reaction's reactants raised to its respective stoichiometric coeff. The equation is Kc=[SO2][Cl2]/[SO2Cl2]= [1.3*10^-2][1.3*10^-2]/[2.2*10^-2-<span>1.3*10^-2]=0.0188. The final answer is Kc=0.0188.</span>
2Ag2S + 2H2O—>4Ag+2H2S+O2
The reactants are silver sulphide (Ag2s) and water (H2O)
