Consider the following half-cell reactions: $\text{Mn}^{2+} + 2e^{-} \rightarrow \text{Mn}$, $E^{o} = -1.18$ V $\text{Mn}^{2+} \rightarrow \text{Mn}^{3+} + e^{-}$, $E^{o} = -1.51$ V The $E^{o}$ for the reaction $3\text{Mn}^{2+} \rightarrow \text{Mn} + 2\text{Mn}^{3+}$, and the possibility of the forward reaction, are respectively: (a) 2.69 V and spontaneous (b) -2.69 V and non-spontaneous (c) 0.33 V and spontaneous (d) 4.18 V and non-spontaneous

Consider the following half-cell reactions: Mn²⁺ + 2e⁻ arrow Mn, Eo = -1.18 V Mn²⁺ arrow Mn³⁺ + e⁻, Eo = -1.51

Choose the Best Answer · Q2

\text{Mn}^{2+} + 2e^{-} \rightarrow \text{Mn}

(a) 2.69 V and spontaneous

(b) -2.69 V and non-spontaneous

(d) 4.18 V and non-spontaneous

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✓ Free question

Concept understanding — Corrosion and the Electrochemical Series

Corrosion — the rusting of iron, or the green film on copper and brass — is a redox process in which a metal is oxidised by atmospheric oxygen in the presence of moisture. Rusting is specifically an electrochemical process requiring a spatially separated anode and cathode on the SAME piece of metal: a water droplet on iron creates a low-oxygen anodic region beneath it, where iron dissolves ( $\text{Fe} \rightarrow \text{Fe}^{2+}+2e^-$ , $E^{o}=0.44$ V), and a high-oxygen cathodic region around it, where oxygen is reduced ( $\text{O}_2+4\text{H}^++4e^- \rightarrow 2\text{H}_2\text{O}$ , $E^{o}=1.23$ V); the overall reaction has $E^{o}=1.67$ V, confirming rusting is spontaneous. Fe²⁺ is further oxidised to Fe³⁺, which precipitates as hydrated rust, Fe2O3·xH2O.

Five protection strategies follow directly from blocking one ingredient corrosion needs: painting (blocks oxygen/moisture physically), galvanising (a sacrificial zinc coating corrodes instead of the iron beneath, since zinc is more easily oxidised), cathodic protection (a sacrificial anode of Mg or Zn, connected but not necessarily coating the whole surface, forces the protected metal to act as cathode), passivation (a strong oxidiser like conc. HNO3 forms a tough protective oxide layer, as aluminium's natural Al2O3 layer does), and alloy formation (stainless steel, an Fe–Cr alloy, resists corrosion far better than plain iron).

All of this connects back to the electrochemical series: ranking standard reduction potentials from most positive (strongest oxidising agents, e.g. F2, top of the table) to most negative (strongest reducing agents, e.g. Li, bottom) lets spontaneity, relative oxidising/reducing strength, and corrosion vulnerability all be read off directly — a species higher in the table oxidises the reduced form of any species lower down, and metals lower in the table (more negative $E^{o}$ , more easily oxidised) are inherently more prone to corrosion, exactly why zinc (below iron) can sacrificially protect it.

Step 1. The target reaction, $3\text{Mn}^{2+} \rightarrow \text{Mn} + 2\text{Mn}^{3+}$ , is a disproportionation: one Mn²⁺ is reduced to Mn(0) while two Mn²⁺ are oxidised to Mn³⁺. Because E° values cannot simply be added when the number of electrons differs between half-reactions, work through Gibbs free energies instead, using $\Delta G = -nFE^{o}$ for each half-reaction as written.

Step 2. Reduction half (n=2): $\text{Mn}^{2+}+2e^- \rightarrow \text{Mn}$ , $E^o=-1.18$ V, so $\Delta G_1 = -(2)(F)(-1.18) = +2.36F$ .

Step 3. Oxidation half, doubled for 2 mol Mn²⁺ (total n=2 electrons released): $2\text{Mn}^{2+} \rightarrow 2\text{Mn}^{3+}+2e^-$ , using the given $E^o=-1.51$ V (this value already IS the potential of the reaction as written, so for 2 electrons $\Delta G_2 = -(2)(F)(-1.51) = +3.02F$ .

Step 4. Total $\Delta G^{o} = \Delta G_1 + \Delta G_2 = 2.36F + 3.02F = 5.38F$ . The overall reaction transfers n = 2 electrons (the 2 electrons released by oxidation are exactly consumed by reduction), so $E^{o}_{overall} = -\dfrac{\Delta G^{o}}{nF} = -\dfrac{5.38F}{2F} = -2.69$ V.

Step 5. Since $E^{o}$ is negative, $\Delta G^{o}$ is positive, so the forward reaction is non-spontaneous — consistent with the chemical fact that Mn²⁺ is the most stable oxidation state of manganese in solution and does not spontaneously disproportionate.

✓Final answer

E^{o} = -2.69

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following statements is incorrect ?$
(a) $Co^{3+}$ and $Fe^{2+}$ have same number of unpaired electrons.
(b) Of all the known elements, Silver has the highest electrical conductivity at room temperature.
(c) To form a substitute alloy, the difference between the atomic radii of solvent and solute should be less than 15%.
(d) If the standard electrode potential ( $E\degree$ ) of a metal is large and negative, the metal is a powerful oxidising agent.
›Reveal solutionSolution
A large negative standard reduction potential signals a strong tendency to lose electrons (be oxidised), which is the defining behaviour of a strong reducing agent, not an oxidising agent — this makes statement (d) the incorrect one among the four options.

Checking each statement:

$Co^{3+}$ ( $d^6$ ) and $Fe^{2+}$ ( $d^6$ ) — both have the same $d^6$ electronic configuration, so under comparable ligand-field conditions they have the same number of unpaired electrons. This statement is correct.

Silver is indeed the element with the highest electrical conductivity of all known metals at room temperature (even higher than copper) — a well-established physical fact. This statement is correct.

The Hume-Rothery size-factor rule for substitutional solid-solution alloys states that if the atomic radii of the solute and solvent metals differ by more than about 15%, solid solubility is very limited; a difference of less than $15\%$ favours extensive substitutional alloy formation. This statement is correct.

The standard electrode potential, $E°$ , is conventionally defined and tabulated as a reduction potential. A metal with a large negative $E°$ has a strong tendency to lose electrons and be oxidised itself (i.e. it is a poor oxidising agent but a powerful reducing agent) — e.g. $Li^+/Li$ , $E° = -3.05\ V$ , and lithium metal is one of the strongest reducing agents known. A powerful oxidising agent, by contrast, is characterised by a large positive $E°$ (a strong tendency to be reduced/gain electrons, e.g. $F_2$ ). So this statement has the relationship backwards, making it incorrect.

✓Final answer
$E°$
CBSE 2019Set ANNUAL1 markMCQ
$Q. The metals near the bottom of the electrochemical series are :$
(a) Weak reducing agents
(b) Weak oxidising agents
(c) Strong oxidising agents
(d) Strong reducing agents
›Reveal solutionSolution

The electrochemical series is arranged by increasing standard reduction potential; metals at the bottom (most positive $E°$ , the noble metals) are the weakest reducing agents because they least readily give up electrons.

Metals at the top of the electrochemical series (very negative $E°$ , e.g. Li, K, Ca, Na) lose electrons most readily and are the strongest reducing agents. Moving down the series, the standard reduction potential becomes progressively more positive, meaning the metal has a decreasing tendency to be oxidised (to lose electrons) and an increasing tendency for its ion to be reduced. Metals near the bottom, such as Cu, Ag and Au ('noble' metals), therefore act as weak reducing agents themselves — it is their corresponding cations that are strong oxidising agents (easily reduced), which is the reverse property to what this question asks about the metals.

✓Final answer
$The correct answer is (a) Weak reducing agents — metals at the bottom of the electrochemical series (noble metals) have little tendency to lose electrons.$

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