Exercises — Galvanic (voltaic) cells — anode (oxidation), cathode (reduction)

What we do: recall the two definitions, no arithmetic.

• Oxidation = loss of electrons. The electrode where oxidation happens is the anode ("An Ox").
• Zinc is the one that loses electrons, so Zn is the anode.
• In a galvanic cell the anode pushes electrons OUT into the wire, so it is the negative terminal.

Answer: Oxidation at the anode (Zn), which is the negative terminal.

Electrons are produced by oxidation at the anode and consumed by reduction at the cathode. So they always flow anode → cathode in the outside wire. Answer: electrons flow Ni → Ag. (Careful: conventional current — the old fluid-arrow convention — points the opposite way, Ag→Ni. We are tracking electrons.)

Step 1 — who gets reduced? The species with the more positive reduction potential wins the electrons. $+0.80>-0.25$, so Ag⁺ is reduced → Ag is the cathode; Ni is oxidized → anode. Step 2 — plug in (both numbers are reduction potentials): $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=0.80-(-0.25)=1.05\text{V}$ Step 3 — sanity: positive $E{°}_{\text{cell}}$ ⇒ spontaneous ⇒ it's a real galvanic cell. ✓ Answer: anode Ni, cathode Ag, $E{°}_{\text{cell}}=1.05\mathbf{V}$.

Why balance electrons first? When you add half-reactions the electrons must cancel exactly, otherwise charge isn't conserved. Ni gives 2 electrons; Ag accepts only 1. Multiply the silver half by 2:

• Anode: $\text{Ni}\to {\text{Ni}}^{2+}+2e^{-}$
• Cathode ($\times 2$): $2{\text{Ag}}^{+}+2e^{-}\to 2\text{Ag}$

Add (the $2e^{-}$ cancel): $\text{Ni}+2{\text{Ag}}^{+}\to {\text{Ni}}^{2+}+2\text{Ag}$ Note: $E{°}_{\text{cell}}$ does not change when you scale a half-reaction — potential is energy per electron, an intensive quantity, so $1.05\text{V}$ still stands.

Step 1: more negative $E°$ = more eager to give electrons = anode. $-0.14<-0.13$, so Sn is the anode, Pb the cathode. Step 2: $E{°}_{\text{cell}}=E{°}_{\text{Pb}}-E{°}_{\text{Sn}}=(-0.13)-(-0.14)=+0.01\text{V}$ Step 3 — meaning: positive, so it works, but the two metals almost equally want the electrons, so the "electron-pressure difference" is tiny — a very weak push. Real currents from such a cell are minuscule and easily reversed by concentration changes (that's where the Nernst Equation takes over). Answer: Sn anode, $E{°}_{\text{cell}}=+0.01\text{V}$ (spontaneous but very weak).

Read the ladder: the vertical axis is $E°$. Lower on the ladder = more negative = more eager to be oxidized = stronger reducing agent.

• Ranking (strongest→weakest reducer): $\text{Zn}(-0.76)>\text{Fe}(-0.44)>\text{Cu}(+0.34)>\text{Ag}(+0.80)$.
• The biggest voltage comes from the pair furthest apart on the ladder: bottom (Zn, anode) with top (Ag, cathode): $E{°}_{\text{cell}}=0.80-(-0.76)=1.56\text{V}$ Answer: Zn > Fe > Cu > Ag as reducers; largest cell = Zn–Ag, $1.56\text{V}$.

Why charge first? Electrons are counted through charge $Q$. This is the bridge to Faraday's Laws of Electrolysis. Step 1 — total charge: $Q=It=0.50\text{A}\times (60\times 60\text{s})=1800\text{C}$. Step 2 — moles of electrons: one mole of electrons carries $F=96485\text{C}$. $n_{e^{-}}=\frac{Q}{F}=\frac{1800}{96485}=0.01866\text{mol}$ Step 3 — moles of Cu: the half-reaction needs 2 electrons per Cu atom, so $n_{\text{Cu}}=\frac{n_{e^{-}}}{2}=0.009328\text{mol}$ Step 4 — mass: $m=n_{\text{Cu}}\times M_{\text{Cu}}=0.009328\times 63.5=0.592\text{g}$. Answer: (a) $\approx 0.0187\text{mol}$ electrons; (b) $\approx 0.59\mathbf{g}$ of Cu.

Why this formula? Electrical work per coulomb is voltage; total electrical work available is charge $\times$ voltage, and (with a sign) that is the free energy released. See Gibs Free Energy and Cell Potential. $\Delta G°=-nFE{°}_{\text{cell}}=-(2)(96485)(1.10)=-212,267\text{J}\approx -212\text{kJ}$ Sign check: negative ⇒ spontaneous ⇒ consistent with a working galvanic cell. ✓ Answer: $\Delta G°\approx -212\mathbf{kJ}$ (per mole of reaction as written).

Strategy: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$; we hunt for a pair whose gap ≈ 1.10 V, excluding Cu. Available (excluding Cu): Ag $+0.80$, Ni $-0.25$, Fe $-0.44$, Zn $-0.76$, Al $-1.66$. Scan cathode = Ag (most positive, best cathode) against candidate anodes:

• Ag–Ni: $0.80-(-0.25)=1.05$
• Ag–Fe: $0.80-(-0.44)=1.24$
• Ag–Zn: $0.80-(-0.76)=1.56$

$1.05\text{V}$ (Ag–Ni) is the closest to $1.10$ (off by $0.05$; Ag–Fe is off by $0.14$). Design:

• Anode (oxidation): $\text{Ni}\to {\text{Ni}}^{2+}+2e^{-}$
• Cathode ($\times 2$): $2{\text{Ag}}^{+}+2e^{-}\to 2\text{Ag}$
• Net: $\text{Ni}+2{\text{Ag}}^{+}\to {\text{Ni}}^{2+}+2\text{Ag}$

$E{°}_{\text{cell}}=0.80-(-0.25)=1.05\mathbf{V}$ Answer: Ni anode, Ag cathode, $1.05\text{V}$ — the closest achievable to $1.10\text{V}$ without copper.

(a) Fe $E°=-0.44$ (more negative) → anode; Ag $E°=+0.80$ → cathode. (b) Fe gives 2 e⁻, Ag takes 1 → double the silver half: $\text{Fe}+2{\text{Ag}}^{+}\to {\text{Fe}}^{2+}+2\text{Ag}(n=2)$ (c) $E{°}_{\text{cell}}=0.80-(-0.44)=1.24\text{V}$. (d) $\Delta G°=-nFE°=-(2)(96485)(1.24)=-239,283\text{J}\approx -239\text{kJ}$. (e) The anode compartment builds up positive ${\text{Fe}}^{2+}$; to neutralise it, anions in the salt bridge migrate toward the anode (Fe side). Cations drift toward the cathode. This is exactly the neutrality job the salt bridge does in the parent note. Answer: Fe anode / Ag cathode; net $\text{Fe}+2{\text{Ag}}^{+}\to {\text{Fe}}^{2+}+2\text{Ag}$; $E{°}_{\text{cell}}=1.24\mathbf{V}$; $\Delta G°\approx -239\mathbf{kJ}$; anions → anode.

Why we can't just reuse $E°$: the ° means "standard, all $1\text{M}$". Change a concentration and the driving force changes. The tool for that is the Nernst Equation, which at $25{}^{\circ }\text{C}$ reads $E_{\text{cell}}=E{°}_{\text{cell}}-\frac{0.0592}{n}{\log }_{10}Q,Q=\frac{[{\text{Zn}}^{2+}]}{[{\text{Cu}}^{2+}]}.$ What $Q$ measures: the "reaction quotient" — how far the products have built up versus reactants right now. Here products (Zn²⁺) are plentiful and reactant (Cu²⁺) is scarce, so $Q$ is large and the push should drop. Step 1 — compute $Q$: $Q=\frac{1.0}{0.010}=100$, so ${\log }_{10}Q=2$. Step 2 — plug in ($n=2$): $E_{\text{cell}}=1.10-\frac{0.0592}{2}(2)=1.10-0.0592=1.041\text{V}.$ Step 3 — direction check: starving the cathode of Cu²⁺ lowers the voltage — exactly what intuition said. If instead Cu²⁺ were more concentrated than $1\text{M}$, $Q<1$, $\log Q<0$, and the voltage would rise above $1.10\text{V}$. Answer: voltage drops by about $0.059\text{V}$ to $\approx 1.04\mathbf{V}$.

Anode reaction type ::: Oxidation (loss of electrons) Sign of the anode in a galvanic cell ::: Negative Formula for standard cell potential ::: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$ Does doubling a half-reaction change $E°$ ::: No — potential is per-electron (intensive) What tool handles non-standard concentrations ::: The Nernst equation, $E=E°-\frac{0.0592}{n}\log Q$