SCH4U Chemistry Practice Exam — Free

Start by deciding which half-reaction runs as a reduction. In a galvanic cell the half-reaction with the more positive standard reduction potential occurs at the cathode, because that species is the stronger oxidizing agent. $\mathrm{Ag^+}$ ($E^\circ = +0.80\ \text{V}$) is a much stronger oxidizing agent than $\mathrm{Ni^{2+}}$ ($E^\circ = -0.26\ \text{V}$), so silver ions are reduced at the silver electrode (the cathode) and nickel metal is oxidized at the nickel electrode (the anode). Oxidation always takes place at the anode, whichever metal it is made of.

The silver half-reaction had to be doubled so that the electrons cancel, but its potential is not doubled. $E^\circ$ is an intensive property (energy per coulomb of charge), so it does not depend on how many moles react. The cell potential is the cathode's reduction potential minus the anode's reduction potential:

A positive cell potential confirms that the reaction is spontaneous as written: electrons flow through the external wire from the nickel anode to the silver cathode, the nickel electrode slowly loses mass, and silver plates onto the silver electrode. In cell notation, $\mathrm{Ni}(s)\,|\,\mathrm{Ni^{2+}}(aq)\,\|\,\mathrm{Ag^+}(aq)\,|\,\mathrm{Ag}(s)$.

Where the distractors come from: option B ($+0.54\ \text{V}$) drops the negative sign on the nickel potential and computes $0.80-0.26$; subtracting a negative potential adds. Option C ($+1.86\ \text{V}$) doubles the silver potential along with its half-reaction, the most common trap in this type of question. Option D has the right voltage but the electrodes backwards: the anode is where oxidation happens, and it is nickel, the metal with the more negative reduction potential, that is oxidized.

$E^\circ_{\text{cell}} = +1.06\ \text{V}$, and the nickel electrode is the anode (option A).

Hess's law says that because enthalpy is a state function, $\Delta H$ for the target reaction equals the sum of the $\Delta H$ values of any set of equations that adds up to it. Rearrange each given equation so that its species end up on the correct side with the correct coefficients, and do the same thing to its $\Delta H$.

The target needs $2\,\mathrm{C}(s)$ on the left, so multiply equation (1) by 2. It needs $3\,\mathrm{H_2}(g)$ on the left, so multiply equation (2) by 3. It needs one $\mathrm{C_2H_6}(g)$ on the right, but equation (3) has two on the left, so reverse equation (3) and divide it by 2. Reversing changes the sign of $\Delta H$; halving changes its size.

Adding the three equations, $2\,\mathrm{CO_2}$ and $3\,\mathrm{H_2O}$ appear on both sides and cancel, and so does oxygen: $2+\tfrac{3}{2}=\tfrac{7}{2}$ moles on the left against $\tfrac{7}{2}$ on the right. What remains is exactly the target equation, so

Check by a second route. Equations (1) and (2) are the formation reactions of $\mathrm{CO_2}(g)$ and $\mathrm{H_2O}(l)$, so for the combustion of one mole of ethane ($\Delta H^\circ = -1560.4\ \text{kJ}$), the formation-enthalpy equation $\Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$ gives $-1560.4 = 2(-393.5) + 3(-285.8) - \Delta H_f^\circ(\mathrm{C_2H_6})$, so $\Delta H_f^\circ(\mathrm{C_2H_6}) = -1644.4 + 1560.4 = -84.0\ \text{kJ/mol}$, the same result (tables list about $-84\ \text{kJ/mol}$).

The common trap is equation (3): forgetting to flip the sign when reversing it gives $-3204.8\ \text{kJ}$, and reversing it without halving gives $+1476.4\ \text{kJ}$. Every operation on an equation must be applied to its $\Delta H$ as well.

$\Delta H^\circ = -84.0\ \text{kJ}$ per mole of ethane formed; the formation of ethane is exothermic.

(a) Methanoic acid is a weak acid, so only a small fraction of its molecules ionize and an equilibrium is established. Water is a pure liquid and is left out of the expression:

Let $x$ be the concentration of acid that ionizes (all values in mol/L). The $1.0\times10^{-7}$ mol/L of $\mathrm{H_3O^+}$ already present from water is negligible.

Because $K_a$ is small, assume $x$ is negligible next to 0.45, so $0.45-x \approx 0.45$:

Validity check (the 5% rule): $x$ is $\dfrac{9.0\times10^{-3}}{0.45}\times100\% = 2.0\%$ of the initial concentration, which is below 5%, so the approximation is acceptable. (Solving the quadratic exactly gives $x = 8.9\times10^{-3}$ mol/L and the same pH to two decimal places.)

The trap here is treating the acid as strong: $-\log(0.45) = 0.35$ is far too acidic, because only 2% of the molecules are actually ionized.

(b) Percent ionization is the fraction ionized at equilibrium:

Diluting the solution lowers the concentration of every dissolved species. By Le Châtelier's principle the equilibrium shifts toward the side with more moles of dissolved particles, the right side (two ions from one molecule), to partly counteract the drop in concentration. Equivalently, $Q = \dfrac{[\mathrm{H_3O^+}][\mathrm{HCOO^-}]}{[\mathrm{HCOOH}]}$ falls below $K_a$ on dilution because the numerator has two concentration factors and the denominator only one, so the forward reaction proceeds. More of the acid ionizes, and the percent ionization increases (at one-tenth the concentration about 6% is ionized), even though $[\mathrm{H_3O^+}]$ itself falls and the pH rises.

(c) Sodium methanoate is a soluble ionic compound: $\mathrm{HCOONa}(aq) \rightarrow \mathrm{Na^+}(aq) + \mathrm{HCOO^-}(aq)$. $\mathrm{Na^+}$ is a spectator ion, but $\mathrm{HCOO^-}$ is the conjugate base of a weak acid, so it accepts protons from water and the solution is basic:

With the same ICE reasoning, $[\mathrm{HCOO^-}] = 0.45-y \approx 0.45$ and $[\mathrm{OH^-}] = y$ (carrying the unrounded $K_b = 5.56\times10^{-11}$):

The approximation is easily valid ($y$ is about 0.001% of 0.45). Then

A pH above 7 is expected for the salt of a weak acid and a strong base. The common mistakes are using $K_a$ instead of $K_b$ for the anion, or stopping at 5.30 and reporting the pOH as the pH.

(a) pH = 2.05; (b) 2.0% ionized, and the percent ionization increases on dilution; (c) pH = 8.70.