Q.Use the data of Q.8 ($E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V$; $E^\circ_{Cl_2/Cl^-} = 1.36\ V$; $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ V$; $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$) and find out the most stable oxidised species.

The key idea is that the standard Gibbs free energy change for a redox reaction is directly related to the standard cell potential by $\Delta G^\circ = -nFE^\circ_\text{cell}$. A positive cell potential means the reaction is spontaneous (negative $\Delta G^\circ$), and a negative cell potential means it is non-spontaneous (positive $\Delta G^\circ$). Here, we are given two half-cell reduction potentials; we must combine them correctly to get the cell potential for the reaction as written.

1. Identify the half-reactions and their standard reduction potentials.

   • Cadmium: $\mathrm{Cd^{2+} + 2e^- \rightarrow Cd(s)}$, $E^\circ = -0.403\ \text{V}$
   • Zinc: $\mathrm{Zn^{2+} + 2e^- \rightarrow Zn(s)}$, $E^\circ = -0.763\ \text{V}$

2. Determine which half-reaction is oxidation and which is reduction in the given overall reaction.

   The reaction is: $\mathrm{Cd^{2+}(aq) + Zn(s) \rightarrow Zn^{2+}(aq) + Cd(s)}$.

   • $\mathrm{Zn(s)}$ loses electrons to become $\mathrm{Zn^{2+}}$: this is oxidation.
   • $\mathrm{Cd^{2+}}$ gains electrons to become $\mathrm{Cd(s)}$: this is reduction. So the cell is: $\mathrm{Zn(s) | Zn^{2+}(aq) || Cd^{2+}(aq) | Cd(s)}$.

3. Calculate the standard cell potential $E^\circ_\text{cell}$.

   The standard cell potential is:

Here, the cathode (reduction) is $\mathrm{Cd^{2+}/Cd}$ and the anode (oxidation) is $\mathrm{Zn^{2+}/Zn}$.

4. Determine the number of electrons transferred ($n$). Both half-reactions involve 2 electrons: $\mathrm{Cd^{2+} + 2e^- \rightarrow Cd}$ and $\mathrm{Zn \rightarrow Zn^{2+} + 2e^-}$. So $n = 2$. …

The key concept here is the relationship between reduction potential and reducing strength. A reducing agent is a species that donates electrons (gets oxidized). The more easily it donates electrons, the stronger it is as a reducing agent. In electrochemistry, the tendency to be oxidized is the opposite of the tendency to be reduced. So, a very negative reduction potential means the species is very hard to reduce — which means its oxidized form is very stable, and the reduced form (the species itself) is very eager to give up electrons. Thus, the more negative the standard reduction potential, the stronger the reducing agent.

Let’s work through the data step by step.

1. List the given half-reactions and their standard reduction potentials (E°)

   • $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$  $E^\circ = +1.33\ \mathrm{V}$
   • $\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$  $E^\circ = +1.51\ \mathrm{V}$
   • $\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-$  $E^\circ = +1.36\ \mathrm{V}$
   • $\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr}$  $E^\circ = -0.74\ \mathrm{V}$

2. Identify the species that are candidates as reducing agents

   The question asks for the strongest reducing agent among the options:

   (A) Cr (B) Mn²⁺ (C) Cr³⁺ (D) Cl⁻

   These are the reduced forms of the couples given. For each, we look at the reduction potential of the couple where that species appears on the right (as the product of reduction).

   • For Cr(s): the couple is $\mathrm{Cr}^{3+}/\mathrm{Cr}$ with $E^\circ = -0.74\ \mathrm{V}$.
   • For Mn²⁺: the couple is $\mathrm{MnO}_4^-/\mathrm{Mn}^{2+}$ with $E^\circ = +1.51\ \mathrm{V}$.
   • For Cr³⁺: the couple is $\mathrm{Cr}_2\mathrm{O}_7^{2-}/\mathrm{Cr}^{3+}$ with $E^\circ = +1.33\ \mathrm{V}$.
   • For Cl⁻: the couple is $\mathrm{Cl}_2/\mathrm{Cl}^-$ with $E^\circ = +1.36\ \mathrm{V}$.

3. Compare reducing strength using the reduction potentials

   The more negative the reduction potential, the stronger the reducing agent. Here:

   • Cr: $E^\circ = -0.74\ \mathrm{V}$ (negative → strong reducing agent) …

Concept

A galvanic cell's standard potential is $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ (both as reduction potentials). Here we must first build $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})$ from the given couples, and potentials are combined through Gibbs energy, not added directly.

Solution

1. Half-reactions. Reduction (cathode): $\mathrm{Ag^+ + e^- \to Ag}$, $E^\circ = a$. Oxidation (anode): $\mathrm{Fe^{2+} \to Fe^{3+} + e^-}$, i.e. the $\mathrm{Fe^{3+}/Fe^{2+}}$ couple.

2. Derive $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})$. With $\Delta G^\circ = -nFE^\circ$:

Subtracting gives $\mathrm{Fe^{3+}+e^-\to Fe^{2+}}$: …

Step 1 — Set up the target reaction from the given half-reactions

We are given:

We need $E°$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$, obtained by adding the first equation to twice the reverse of the second equation:

Adding gives $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$, with the $2e^-$ cancelling — an overall $n = 2$ electron process.

Step 2 — Convert each step to $\Delta G°$ (never add $E°$ values directly)

Reversing the second half-reaction flips the sign of $E°_2$; doubling it doubles $\Delta G°$ (since $\Delta G°$, unlike $E°$, is extensive):

$\Delta G°_{2,\ \text{reversed} \times 2} = -(2)F(-1.51) = +3.02F$ …

Step 1 — Fix the four definitions

• Combination redox: two substances combine to give one product, with an oxidation-number change. General form $\ce{A + B -> AB}$.
• Decomposition redox: a single compound breaks into two or more products, with an oxidation-number change. General form $\ce{AB -> A + B}$.
• Displacement redox: a more reactive element displaces a less reactive one from its compound. General form $\ce{X + YZ -> XZ + Y}$.
• Disproportionation: one element in one substance is simultaneously oxidised and reduced. This requires the element to have an intermediate oxidation state, so it has somewhere to go both up and down.

Step 2 — Example (iii): $\ce{CH4_{(g)} + 2O2_{(g)} ->[\Delta] CO2_{(g)} + 2H2O_{(l)}}$

Carbon: $-4$ in $\ce{CH4}$ $\to$ $+4$ in $\ce{CO2}$ (oxidised).

Oxygen: $0$ in $\ce{O2}$ $\to$ $-2$ in $\ce{CO2}$ and $\ce{H2O}$ (reduced).

Methane and oxygen combine into the products, so this is a combination reaction. $\;\boxed{\text{a} \to \text{iii}}$

Step 3 — Example (iv): $\ce{2H2O_{(l)} ->[\text{electrolysis}] 2H2_{(g)} + O2_{(g)}}$

Hydrogen: $+1 \to 0$ (reduced). Oxygen: $-2 \to 0$ (oxidised).

A single compound splits into two elements — a textbook decomposition reaction. $\;\boxed{\text{b} \to \text{iv}}$

Step 4 — Example (i): $\ce{Cl2_{(g)} + 2Br^-_{(aq)} -> 2Cl^-_{(aq)} + Br2_{(l)}}$

Chlorine: $0 \to -1$ (reduced). Bromine: $-1 \to 0$ (oxidised). …

The trap here is that electrode potentials are not additive like voltages in series. Because they are intensive properties (energy per charge), you must convert each potential to a Gibbs free energy change ($\Delta G^\circ = -nFE^\circ$), sum the free energies, and then convert back to a potential for the desired half‑reaction.

Step‑by‑step reasoning

1. Write the given half‑reactions and their standard potentials

   • (1) $\mathrm{Cu}^+ + e^- \rightarrow \mathrm{Cu}$ $E^\circ_1 = +0.50\ \mathrm{V}$, $n_1 = 1$
   • (2) $\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}$ $E^\circ_2 = +0.335\ \mathrm{V}$, $n_2 = 2$

2. Convert each to Gibbs free energy

   Use $\Delta G^\circ = -nFE^\circ$ (with $F$ the Faraday constant; it cancels later).

   • For (1): $\Delta G^\circ_1 = -1 \cdot F \cdot 0.50 = -0.50F$
   • For (2): $\Delta G^\circ_2 = -2 \cdot F \cdot 0.335 = -0.67F$

3. Find the free energy for the target half‑reaction

   Target: $\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^+$

   This can be obtained by reversing reaction (1) and adding it to reaction (2):

   • Reverse (1): $\mathrm{Cu} \rightarrow \mathrm{Cu}^+ + e^-$ gives $\Delta G^\circ = +0.50F$
   • Add to (2): $\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}$ gives $\Delta G^\circ = -0.67F$
   • Sum: $\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^+$ …

Concept & Intuition

The standard reduction potential $E^\circ$ measures how easily a species gains electrons (is reduced). A higher $E^\circ$ means a stronger oxidising agent — it readily takes electrons. Permanganate in acid, $\mathrm{MnO}_4^-/\mathrm{H}^+$, has $E^\circ = +1.49\ \mathrm{V}$. For it to oxidise a metal ion, that metal ion’s reduction potential must be lower than 1.49 V. Why? Because the metal ion would need to lose electrons (be oxidised), and that is the reverse of its reduction half-reaction. The reverse reaction’s tendency is given by $-E^\circ$. If the metal ion’s $E^\circ$ is greater than permanganate’s, then the metal ion is actually a stronger oxidant than permanganate — it won’t give up electrons to permanganate; instead, permanganate would be the one reduced. So we simply check which metals have $E^\circ > +1.49\ \mathrm{V}$.

Step-by-step reasoning

1. List the given reduction potentials

   • $\mathrm{MnO}_4^-/\mathrm{H}^+$: $E^\circ = +1.49\ \mathrm{V}$
   • (a) $\mathrm{Co}^{3+}/\mathrm{Co}^{2+}$: $+1.81\ \mathrm{V}$
   • (b) $\mathrm{Cr}^{3+}/\mathrm{Cr}$: $-0.74\ \mathrm{V}$
   • (c) $\mathrm{Au}^{3+}/\mathrm{Au}$: $+1.50\ \mathrm{V}$
   • (d) $\mathrm{Ag}^+/\mathrm{Ag}$: $+0.80\ \mathrm{V}$

2. Compare each to +1.49 V

   • For a metal ion to be oxidised by permanganate, its reduction potential must be less than 1.49 V.
   • (a) $1.81 > 1.49$ → cannot be oxidised. …

The couples are given as $M/M^{n+}$ with these values (oxidation potentials):

• [A] $\mathrm{Cu/Cu^{2+}}$: $-0.34$ V
• [B] $\mathrm{Ag/Ag^{+}}$: $-0.8$ V
• [C] $\mathrm{Ca/Ca^{2+}}$: $+2.87$ V
• [D] $\mathrm{Cr/Cr^{3+}}$: $+0.74$ V

A stronger reducing agent is more easily oxidised, i.e. has the more positive oxidation potential. Arranging in increasing order of these values: …

Step 1 — The observation to explain.

$\mathrm{CuF_2}$, $\mathrm{CuCl_2}$ and $\mathrm{CuBr_2}$ all exist, but $\mathrm{CuI_2}$ does not. So the question is: what is special about iodide?

Step 2 — The redox answer.

Iodide is the most easily oxidised halide (largest, most polarisable, lowest ionisation energy for the extra electron; $E^{\circ}_{\mathrm{I_2/I^-}}=+0.54\ \mathrm{V}$, the least positive of the halogen couples). Copper(II) is a mild oxidant that becomes a strong one when the product $\mathrm{Cu^{+}}$ can be locked away as an insoluble salt — and CuI is exactly that. So the moment you try to make $\mathrm{CuI_2}$, the ions react:

$\mathrm{Cu^{2+}}$ is reduced to $\mathrm{Cu^{+}}$ while $\mathrm{I^{-}}$ is oxidised to $\mathrm{I_2}$. The driving force is twofold: (i) the very favourable lattice/solubility term — CuI is highly insoluble, which pulls the equilibrium over; and (ii) iodide's low oxidation potential. With $\mathrm{F^-,Cl^-,Br^-}$ this electron transfer is not favourable enough, so their Cu(II) halides survive.

This is the same redox pair used in the classic iodometric estimation of copper, where the liberated $\mathrm{I_2}$ is titrated with hypo ($\mathrm{Na_2S_2O_3}$) — proof that the reaction really runs.

Step 3 — Reject the others. …

Step 1 — Find $n$ from the balanced cell reaction.

Oxidation: $\mathrm{H_2 \rightarrow 2H^+ + 2e^-}$

Reduction: $2\mathrm{AgCl} + 2e^- \rightarrow 2\mathrm{Ag} + 2\mathrm{Cl^-}$

Electrons transferred: $n = 2$.

Step 2 — The bridge between $E^\circ$ and $K$.

At equilibrium the cell is dead ($E_{\text{cell}} = 0$, $Q = K$), so the Nernst equation

gives

(Equivalently from $\Delta G^\circ = -nFE^\circ = -RT\ln K$ at $T = 298\,$K.)

Step 3 — Substitute.

Step 4 — Take the antilog. …

Reasoning. The reduction potential measures how easily a species accepts electrons (is reduced). The larger the value, the stronger the oxidising agent.

$E^\circ_X=1.78\,\text{V},\qquad E^\circ_Y=1.09\,\text{V},\qquad E^\circ_X>E^\circ_Y$ …