Exercises — Standard electrode potentials — SHE reference, electrochemical series

What we ask: an oxidizing agent takes electrons (gets reduced). Stronger = wants electrons more = more positive $E°$.

• ${\text{Zn}}^{2+}:-0.76$, ${\text{Cu}}^{2+}:+0.34$, ${\text{Cl}}_2:+1.36$. Highest is ${\text{Cl}}_2$.
• Strongest oxidizing agent = ${\text{Cl}}_2$.

A reducing agent gives electrons (gets oxidized). Stronger = wants to keep electrons less = its reduction $E°$ is more negative. Among $\text{Zn}(-0.76),\text{Cu}(+0.34),\text{Ag}(+0.80)$ metals, the most-negative belongs to $\text{Zn}$.

• Strongest reducing agent = $\text{Zn}$.

$E{°}_{{\text{H}}^{+}/{\text{H}}_2}=0.00$ and $E{°}_{{\text{Cu}}^{2+}/\text{Cu}}=+0.34$. Since $0.00<0.34$, hydrogen sits above copper (table) / to the left of copper (figure). Physically: ${\text{Cu}}^{2+}$ pulls electrons harder than ${\text{H}}^{+}$ does. So ${\text{Cu}}^{2+}$ can be reduced by ${\text{H}}_2$, but ${\text{H}}^{+}$ cannot be reduced by $\text{Cu}$. This is exactly why copper does not dissolve in dilute non-oxidizing acid to give ${\text{H}}_2$.

Read the diagram: SHE is on the left (anode), M is on the right (cathode) — and indeed metal deposits ⇒ reduction at the M electrode ⇒ M is the cathode. Both values entered as reduction potentials: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=E{°}_{{\text{M}}^{2+}/\text{M}}-0.00$ $+0.34=E{°}_{{\text{M}}^{2+}/\text{M}}⟹\boxed{E{°}_{{\text{M}}^{2+}/\text{M}}=+0.34\text{V}}$ (This is copper — see the table.)

Use the diagram convention: zinc on the left = anode, silver on the right = cathode. (This matches the table too: higher $E°$ reduces, and ${\text{Ag}}^{+}(+0.80)>{\text{Zn}}^{2+}(-0.76)$.) Both entered as reduction potentials: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=(+0.80)-(-0.76)=+1.56\text{V}$ Positive ⇒ this cell drives current spontaneously as written.

Key idea: $E°$ is an intensive property — a per-coulomb voltage. Multiplying a half-reaction (to balance electrons) does not scale $E°$. Only $\Delta G°$ (extensive) scales with $n$. Cathode ${\text{Ag}}^{+}/\text{Ag}=+0.80$, anode ${\text{Cu}}^{2+}/\text{Cu}=+0.34$ (both reduction potentials): $E{°}_{\text{cell}}=0.80-0.34=+0.46\text{V}$ Even though we used $2{\text{Ag}}^{+}$, we still use $E{°}_{{\text{Ag}}^{+}/\text{Ag}}=+0.80\text{V}$, unchanged.

Proposed: ${\text{Br}}_2+2{\text{I}}^{-}\to 2{\text{Br}}^{-}+{\text{I}}_2$.

• Reduction (cathode): ${\text{Br}}_2+2e^{-}\to 2{\text{Br}}^{-},E°=+1.07$
• Oxidation (anode): $2{\text{I}}^{-}\to {\text{I}}_2+2e^{-}$, from $E{°}_{{\text{I}}_2/{\text{I}}^{-}}=+0.54$ Both entered as reduction potentials (do not flip the $+0.54$): $E{°}_{\text{cell}}=1.07-0.54=+0.53\text{V}>0$ Spontaneous. ${\text{Br}}_2$ (higher $E°$) is the stronger oxidizer and wins, forcing ${\text{I}}^{-}$ to oxidize.

Now cathode would be ${\text{I}}_2/{\text{I}}^{-}(+0.54)$, anode ${\text{Br}}_2/{\text{Br}}^{-}(+1.07)$: $E{°}_{\text{cell}}=0.54-1.07=-0.53\text{V}<0$ Non-spontaneous on its own (though drivable electrolytically with an external voltage $>0.53\text{V}$). A redox pair can only run one way spontaneously; the reverse of a favourable reaction is unfavourable by the same magnitude. That is why 3.1 and 3.2 are exact negatives of each other.

Step 1 — pick the two half-reactions (as reductions):

• Cathode: ${\text{Fe}}^{3+}+e^{-}\to {\text{Fe}}^{2+}$, $E°=+0.77$ — carries 1 electron.
• Anode (will be reversed): ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$, $E°=+0.34$ — carries 2 electrons.

Step 2 — balance the electrons. Copper releases 2 electrons but each iron only accepts 1. To make the electrons given equal the electrons taken, multiply the iron half-reaction by 2: $2{\text{Fe}}^{3+}+2e^{-}\to 2{\text{Fe}}^{2+}(\text{now 2 electrons, matching Cu})$ $\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}(\text{oxidation, 2 electrons})$ Adding them, the $2e^{-}$ cancel cleanly: $2{\text{Fe}}^{3+}+\text{Cu}\to 2{\text{Fe}}^{2+}+{\text{Cu}}^{2+},n=2.$

Step 3 — compute $E{°}_{\text{cell}}$ (note $E°$ is unchanged by the factor of 2): $E{°}_{\text{cell}}=0.77-0.34=+0.43\text{V}>0$ Yes, spontaneous. The $n=2$ we found here is exactly the $n$ you would feed into $\Delta G°=-nFE{°}_{\text{cell}}$.

First check $n$: the balanced reaction is $\text{Zn}+{\text{Cu}}^{2+}\to {\text{Zn}}^{2+}+\text{Cu}$; both half-reactions already carry 2 electrons, so $n=2$ (total electrons in the overall reaction). $\Delta G°=-nFE{°}_{\text{cell}}=-(2)(96485)(1.10)\text{J}=-212267\text{J}\approx -212.3\text{kJ}$ Negative ⇒ spontaneous, consistent with $E{°}_{\text{cell}}>0$. This is the numerical bridge to Gibbs energy and $K$.

Where does $0.0592$ come from? Start from two facts you already have: $\Delta G°=-nFE{°}_{\text{cell}}$ and (from thermodynamics) $\Delta G°=-RT\ln K$. Setting them equal and solving: $nFE{°}_{\text{cell}}=RT\ln K⟹{\log }_{10}K=\frac{nFE{°}_{\text{cell}}}{2.303RT}.$ Plug in $R=8.314$, $T=298\text{K}$, $F=96485$: $\frac{2.303RT}{F}=\frac{2.303\times 8.314\times 298}{96485}=0.0592\text{V}.$ So $0.0592\text{V}$ is simply $2.303RT/F$ at $298\text{K}$ — it is not magic, it is room temperature baked into a constant (same constant powers the Nernst equation).

Now the numbers: ${\log }_{10}K=\frac{(2)(1.10)}{0.0592}=\frac{2.20}{0.0592}=37.16$ $K\approx 10^{37.16}\approx 1.4\times 10^{37}$ A gigantic $K$ ⇒ the Zn/Cu reaction goes essentially to completion — exactly what "spontaneous" predicts.

Step 1 — the sign convention, made explicit. The cell is written $\text{SHE}\Vert \text{metal}$, so by the left-anode/right-cathode rule the metal electrode is treated as the cathode in the diagram. The master equation gives $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=E{°}_{\text{metal}}-E{°}_{\text{SHE}}=E{°}_{\text{metal}}-0=E{°}_{\text{metal}}.$ So with this fixed orientation the measured $E{°}_{\text{cell}}$ equals the metal's reduction potential directly, sign included:

• $E{°}_A=-2.37\text{V}$ (this is $\text{Mg}$),
• $E{°}_B=+0.80\text{V}$ (this is $\text{Ag}$),
• $E{°}_C=-0.44\text{V}$ (this is $\text{Fe}$).

(If a metal's true tendency is to dissolve, the number simply comes out negative — the algebra handles it; we never guess "which is really the anode.")

Step 2 — rank as reducing agents, and WHY by increasing negativity. A reducing agent works by giving electrons away, i.e. by undergoing the reverse of the tabulated reduction. The more negative the reduction $E°$, the less the couple wants electrons, so the more eagerly it hands them off — the stronger the reducer. Sorting from most-negative to least-negative therefore sorts from strongest to weakest reducer: $\underset{\text{strongest reducer}}{\underbrace{A(-2.37)}}>\underbrace{C(-0.44)}>\underset{\text{weakest reducer}}{\underbrace{B(+0.80)}}$ (The "$>$" here means "stronger reducer than," not "larger number" — that is the counter-intuitive part.)

Step 3 — will A displace B's ion? A is the far stronger reducer, so metal A gives electrons to B's ion. Put B's ion as cathode, A as anode (both reduction potentials in the formula): $E{°}_{\text{cell}}=E{°}_B-E{°}_A=0.80-(-2.37)=+3.17\text{V}>0$ Yes — Mg readily displaces ${\text{Ag}}^{+}$, plating out silver. This single-displacement connects straight to cell potential.

• Cathode candidate: ${\text{Fe}}^{3+}/{\text{Fe}}^{2+}(+0.77)$
• Anode candidate: ${\text{Br}}_2/{\text{Br}}^{-}(+1.07)$ $E{°}_{\text{cell}}=0.77-1.07=-0.30\text{V}<0⟹\text{non-spontaneous.}$ ${\text{Fe}}^{3+}$ cannot oxidize ${\text{Br}}^{-}$ on its own. Now test iodide: ${\text{I}}_2/{\text{I}}^{-}(+0.54)$: $E{°}_{\text{cell}}=0.77-0.54=+0.23\text{V}>0⟹\text{spontaneous.}$ ${\text{Fe}}^{3+}$ oxidizes ${\text{I}}^{-}$ but not ${\text{Br}}^{-}$ — because ${\text{I}}^{-}$ sits below ${\text{Fe}}^{3+}$'s reach on the staircase while ${\text{Br}}^{-}$ sits above it.

Figure 2 (below) makes this cutoff literal: it plots the four couples on the $E°$ axis and shades the "reach of ${\text{Fe}}^{3+}$" — every couple to the left of the $+0.77$ line (${\text{I}}^{-}$) gets oxidized, every couple to the right (${\text{Br}}^{-}$, ${\text{Cl}}^{-}$) is safe. Trace the teal bar back to ${\text{I}}_2/{\text{I}}^{-}$ to see exactly why iodide is the only casualty here.