Why does copper not replace hydrogen from acids?

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Q. Why does copper not replace hydrogen from acids?

Telangana TsbieShort· 2mImportance★★★★★

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Concept understanding — Standard Reduction Potential

Cell Representation and the Nernst Equation: From Intuition to Precision

Imagine you have a Daniell cell — a zinc rod in zinc sulphate solution connected by a salt bridge to a copper rod in copper sulphate solution. You know it produces a voltage. But what happens if you dilute the copper sulphate solution? Or if you change the temperature? The voltage changes. The Nernst equation is the tool that tells you exactly how much it changes.

The Intuition First

A battery works because the two half-cells "want" to react — zinc wants to lose electrons, copper ions want to gain them. This "want" is measured as a tendency, or potential. But the strength of that tendency depends on how crowded the ions are.

Think of it like this: If you have a room full of people who all want to leave (like zinc ions wanting to form), the push to get out is stronger when the room is packed. If the room is nearly empty, the push is weaker. Similarly, for copper ions wanting to enter the metal (gain electrons), the pull is stronger when there are many copper ions around, and weaker when there are few.

The Nernst equation quantifies this: the actual cell potential depends on the concentrations (or activities) of the ions involved.

The Precise Statement

For a general cell reaction:

aA + bB \rightarrow cC + dD

The cell potential $E$ under non-standard conditions is given by:

E = E^\circ - \frac{RT}{nF} \ln Q

Where:

$E$ = cell potential under the given conditions (in volts)
$E^\circ$ = standard cell potential (when all reactants/products are at 1 M, 1 atm, 25°C)
$R$ = universal gas constant (8.314 J/mol·K)
$T$ = temperature in Kelvin
$n$ = number of moles of electrons transferred in the balanced half-reactions
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for dilute solutions)

At 25°C (298 K), the equation simplifies to a very practical form:

E = E^\circ - \frac{0.0591}{n} \log_{10} Q

The 0.0591 comes from $\frac{2.303RT}{F}$ at 298 K. Notice it uses $\log_{10}$ (common log), not natural log.

Cell Representation: How We Write It

In electrochemistry, we represent a cell with a shorthand notation. For the Daniell cell:

\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu(s)}

The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double line $\parallel$ represents the salt bridge.

The anode (oxidation) is written on the left, the cathode (reduction) on the right. Electrons flow from left to right in the external circuit.

Applying the Nernst Equation to a Cell Representation

For the Daniell cell, the half-reactions are:

Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
Cathode (reduction): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Overall: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Here $n = 2$ (two electrons transferred). The reaction quotient is:

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

So the Nernst equation becomes:

E = E^\circ - \frac{0.0591}{2} \log_{10} \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Note

Solids (Zn, Cu) do not appear in $Q$ because their concentrations are constant (activity = 1).

A Worked Example

Suppose you have a Daniell cell where $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1.0$ M at 25°C. $E^\circ$ for the cell is 1.10 V. …

Why this formula?

Standard Reduction Potential: Why the Formula Holds

Let's build this from first principles — understanding why before memorising what.

1. The Core Idea: A Half-Cell's "Tendency to Gain Electrons"

A standard reduction potential ( $E^\circ$ ) measures how strongly a species wants to gain electrons (be reduced) under standard conditions (1 M concentration, 1 atm pressure, 25°C).

But why can't we measure this directly? Because every reduction must be paired with an oxidation — you can't have electrons flowing without a complete circuit.

2. The Formula: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

Why subtraction, not addition?

Consider a Daniell cell:

Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu

Experimentally, we measure the cell potential as +1.10 V.

Now, we define the standard hydrogen electrode (SHE) as exactly 0.00 V:

2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad E^\circ = 0.00 \text{ V}

The reasoning step-by-step:

We can only measure differences — like measuring height difference between two points.
If we connect the SHE to the copper half-cell, we measure +0.34 V (Cu²⁺ is reduced).
If we connect the SHE to the zinc half-cell, we measure −0.76 V (Zn²⁺ is reduced less readily than H⁺).

Now, the cell potential is the difference in their tendencies:

E^\circ_{\text{cell}} = E^\circ_{\text{Cu}} - E^\circ_{\text{Zn}} = (+0.34) - (-0.76) = +1.10 \text{ V}

Key insight: The formula uses subtraction because we're comparing two half-cells against the same reference (SHE). The cathode is where reduction happens (higher $E^\circ$ ), the anode is where oxidation happens (lower $E^\circ$ ).

3. The Nernst Equation: Why $E = E^\circ - \frac{RT}{nF} \ln Q$

This is the thermodynamic derivation — the real "why."

From Gibbs free energy:

\Delta G = \Delta G^\circ + RT \ln Q

For an electrochemical cell:

\Delta G = -nFE \quad \text{and} \quad \Delta G^\circ = -nFE^\circ

Substituting:

-nFE = -nFE^\circ + RT \ln Q

Rearranging:

E = E^\circ - \frac{RT}{nF} \ln Q

Why this makes physical sense:

$RT \ln Q$ represents the entropy penalty of non-standard concentrations
$nF$ converts charge to energy (Faraday's constant × number of electrons)
The minus sign means: as products accumulate ( $Q$ increases), the cell potential drops — the reaction is approaching equilibrium

At equilibrium ( $Q = K$ ), $E = 0$ — the battery is "dead."

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The key idea is the Standard Reduction Potential ( $E^\circ$ ) of the copper half-reaction compared to that of hydrogen.

For a metal to displace hydrogen from an acid, the metal must be a stronger reducing agent than hydrogen — meaning its standard reduction potential must be less than $0.00\ \text{V}$ (the $E^\circ$ of $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ ).
The standard reduction potential for copper is:

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s) \quad E^\circ = +0.34\ \text{V}

This positive value means $\text{Cu}^{2+}$ is more easily reduced than $\text{H}^+$ — copper metal is a **weaker** reducing agent than hydrogen gas. …

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The key idea is that a metal can displace hydrogen from an acid only if its standard reduction potential is more negative than $0.00\ \text{V}$ (the standard hydrogen electrode). Copper has $E^\circ = +0.34\ \text{V}$ , which is positive, so it cannot reduce $\ce{H+}$ to $\ce{H2}$ — the reaction is thermodynamically unfavourable.

The Concept: Standard Reduction Potential as a “Tug-of-War” for Electrons

Every metal has an inbuilt tendency to lose electrons and go into solution as positive ions. Chemists measure this tendency using the standard reduction potential ( $E^\circ$ ). Think of it as a ranking: the more negative the $E^\circ$ value, the stronger the metal is as a reducing agent — it wants to give away electrons. The more positive the value, the weaker it is as a reducing agent; it prefers to stay as the metal.

The standard hydrogen electrode (SHE) is the reference point, assigned $E^\circ = 0.00\ \text{V}$ for the half-reaction:

2\ce{H+} + 2e^- \rightarrow \ce{H2}

For a metal to displace hydrogen from an acid, the metal must be able to donate electrons to $\ce{H+}$ ions. That means the metal’s own reduction half-reaction must have a more negative $E^\circ$ than $0.00\ \text{V}$ . Only then will the overall cell potential be positive, making the reaction spontaneous.

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} > 0

Now let’s see where copper stands.

Step-by-Step Reasoning

Write the half-reactions for copper and hydrogen. Copper’s reduction half-reaction is:

\ce{Cu^{2+} + 2e^- -> Cu} \quad E^\circ = +0.34\ \text{V}

Hydrogen’s reduction half-reaction is:

2\ce{H+} + 2e^- \rightarrow \ce{H2} \quad E^\circ = 0.00\ \text{V}

Identify which half-reaction would be the anode (oxidation) and which the cathode (reduction) if copper were to displace hydrogen.

For copper to displace hydrogen, copper metal must be oxidised (lose electrons) and $\ce{H+}$ must be reduced (gain those electrons). So:
- Anode (oxidation): $\ce{Cu -> Cu^{2+} + 2e^-}$ — this is the reverse of the reduction half-reaction, so its potential is $-0.34\ \text{V}$ .
- Cathode (reduction): $2\ce{H+} + 2e^- \rightarrow \ce{H2}$ — potential $0.00\ \text{V}$ .
Calculate the standard cell potential.

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.00\ \text{V} - (-0.34\ \text{V}) = -0.34\ \text{V}

The negative value tells us the reaction is non-spontaneous under standard conditions. In other words, copper does not have the thermodynamic drive to push electrons onto $\ce{H+}$ .

Compare with a metal that does displace hydrogen, like zinc. Zinc has $E^\circ = -0.76\ \text{V}$ $E^{\circ} = - 0.76 V$ for $\ce{Zn^{2+} + 2e^- -> Zn}$ $Zn X^{2 +} + 2 e X^{-} Zn$ . For the same setup:
- Anode (oxidation): $\ce{Zn -> Zn^{2+} + 2e^-}$ , potential $+0.76\ \text{V}$ .
- Cathode (reduction): $2\ce{H+} + 2e^- \rightarrow \ce{H2}$ , potential $0.00\ \text{V}$ . …

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Common Mistakes: Why Copper Doesn't Replace Hydrogen from Acids

The Core Concept First

The question tests your understanding of Standard Reduction Potential ( $E^\circ$ ) and the Electrochemical Series. The short answer: Copper has a positive standard reduction potential ( $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$ ), while hydrogen has $E^\circ_{\text{H}^+/\text{H}_2} = 0.00\ \text{V}$ . For a metal to displace hydrogen from an acid, it must have a more negative reduction potential (i.e., be a stronger reducing agent).

Mistake #1: Confusing Oxidation and Reduction Potentials

The error: Students look at the reduction potential of copper ( $+0.34\ \text{V}$ ) and think "positive means it can reduce hydrogen ions."

Why it's wrong: The reaction we need is:

\text{Cu}(s) + 2\text{H}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + \text{H}_2(g)

Here, copper is oxidized (loses electrons), and $\text{H}^+$ is reduced (gains electrons). The cell potential is:

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.00 - (+0.34) = -0.34\ \text{V}

A negative $E^\circ_{\text{cell}}$ means the reaction is non-spontaneous.

How to avoid: Always write the full redox reaction first. Identify which species is oxidized and which is reduced. Then apply $E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}}$ .

Mistake #2: Memorizing Without Understanding the Series

The error: Students remember "metals above hydrogen in the reactivity series displace hydrogen" but apply it blindly to copper.

Why it's wrong: The reactivity series is based on oxidation tendency. Metals above hydrogen (like Zn, Fe) have more negative reduction potentials and can reduce $\text{H}^+$ . Copper is below hydrogen — it cannot.

How to avoid: Draw the electrochemical series from most negative to most positive:

\text{Li} (-3.04) > \text{Zn} (-0.76) > \text{Fe} (-0.44) > \text{H} (0.00) > \text{Cu} (+0.34) > \text{Ag} (+0.80)

Only metals to the left of hydrogen can displace it from acids.

Mistake #3: Forgetting the Role of Oxidizing Acids

The error: Students say "copper never reacts with acids."

Why it's wrong: Copper does react with oxidizing acids like concentrated $\text{HNO}_3$ or hot concentrated $\text{H}_2\text{SO}_4$ . But here, the $\text{NO}_3^-$ or $\text{SO}_4^{2-}$ ion (not $\text{H}^+$ ) is the oxidizing agent:

\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu}(\text{NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}

How to avoid: Distinguish between non-oxidizing acids (HCl, dilute $\text{H}_2\text{SO}_4$ ) where only $\text{H}^+$ can oxidize, and oxidizing acids where the anion does the oxidation.

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

TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$Q. Copper matte is a mixture of (A) Oxides of Cu and Fe (B) Carbonates of Cu and Fe (C) Sulphides of Cu and Fe (D) Silicates of Cu and Fe$
›Reveal solutionSolution
Copper matte is the intermediate product in copper smelting, consisting mainly of copper and iron sulphides, so the correct answer is (C).

The key concept here is the pyrometallurgical extraction of copper. In the smelting of copper ores (typically chalcopyrite, $\text{CuFeS}_2$ ), the ore is first concentrated by froth flotation, then roasted and smelted in a furnace. The purpose of smelting is to separate the desired metal values from unwanted gangue (rock). The molten product that separates from the slag is called matte.

Why sulphides?

Copper and iron have a strong affinity for sulphur. During smelting, the oxides of these metals (if present) react with remaining sulphur or with added sulphide minerals to form a liquid sulphide phase. This molten sulphide mixture is immiscible with the lighter slag (which contains silicates and oxides). The matte layer sinks to the bottom and is tapped off for further processing (converting to blister copper).

Step-by-step reasoning:

Identify the process context: The question is about "copper matte," which is a specific term from the metallurgy of copper. It is not the final pure copper, but an intermediate product.

Recall the composition of typical copper ores: The most common copper ore is chalcopyrite ( $\text{CuFeS}_2$ ). It contains copper, iron, and sulphur. Other ores like bornite ( $\text{Cu}_5\text{FeS}_4$ ) also contain these three elements.

Understand the smelting reaction: When the concentrated ore is smelted with silica (flux) at high temperature, the iron oxide reacts with silica to form slag (iron silicate), while the copper(I) sulphide and remaining iron(II) sulphide form a separate liquid layer — the matte. The overall reaction can be simplified as:

$\text{CuFeS}_2 + \text{O}_2 + \text{SiO}_2 \rightarrow \text{Cu}_2\text{S} \cdot \text{FeS} \ (\text{matte}) + \text{FeSiO}_3 \ (\text{slag}) + \text{SO}_2$
The matte is essentially a solution of $\text{Cu}_2\text{S}$ and $\text{FeS}$ .

Eliminate other options:

(A) Oxides of Cu and Fe: Oxides are not the main components of matte; they are either reduced or slagged off. Copper oxide would be reduced by sulphide or by added coke. …
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TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
$\ce{CuSO4.2H2O}$
›Reveal solutionSolution
The key idea is that a substance that spontaneously loses its water of crystallisation when exposed to air is called efflorescent. The correct answer is (C).

Concept & Intuition

This question tests your understanding of three common terms describing how solids interact with atmospheric moisture.

Hygroscopic substances absorb moisture from the air but do not necessarily change their physical state (e.g., silica gel).

Deliquescent substances absorb so much moisture that they dissolve in it, forming a solution (e.g., calcium chloride).

Efflorescent substances do the opposite: they lose their own water of crystallisation to the air, often crumbling or becoming powdery (e.g., washing soda, $\ce{Na2CO3.10H2O}$ ).

Isomorphous refers to substances that have the same crystal structure, not a moisture-related property.

The question explicitly says “loses its water of crystallisation on exposure to the atmosphere” — that is the textbook definition of efflorescence.

Step-by-step reasoning

Identify the process described

The substance is losing water that is chemically bound in its crystal structure (water of crystallisation) simply by being in open air. This is a spontaneous dehydration driven by a lower vapour pressure of the hydrated crystal compared to the surrounding air.

Match the process to the correct term

Efflorescence is exactly this: hydrated salts that give up water to the atmosphere.

Hygroscopy involves gaining water from air, not losing it.

Deliquescence is an extreme form of hygroscopy where the substance becomes wet and dissolves.

Isomorphism is unrelated to water loss.

Eliminate the other options

(A) Hygroscopic: Incorrect — these absorb moisture. …
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TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$Q. Lyophilic sols are more stable than lyophobic sols, because (A) The colloidal particles are not solvated (B) There is a strong electrostatic interaction between the colloidal particles (C) The colloidal particles have no change (D) The Brownian movement$
›Reveal solutionSolution
Lyophilic sols are more stable than lyophobic sols primarily because their particles are heavily solvated (surrounded by a solvent layer), which prevents coagulation — the correct option is (B) is not correct; the actual reason is solvation, so the answer is (A) is false, (C) is false, (D) is irrelevant; the key is that lyophilic particles are solvated, giving them stability.

Concept & Intuition

Colloidal stability is about keeping particles from clumping together (coagulating). Lyophilic (“solvent-loving”) sols, like gelatin or starch in water, have particles that strongly attract the solvent molecules. This creates a thick, tightly bound solvent layer around each particle — a “solvation shell.” This shell physically prevents particles from getting close enough to stick. In contrast, lyophobic (“solvent-hating”) sols, like gold or silver in water, lack this shell; their stability depends mainly on electrostatic repulsion from surface charges, which is easier to disrupt (e.g., by adding salt). So the primary reason lyophilic sols are more stable is solvation, not electrostatic interactions or Brownian motion.

Step-by-step reasoning

Understand the terms

Lyophilic: particles have a strong affinity for the dispersion medium (e.g., water). They are often macromolecules or highly hydrated.

Lyophobic: particles have little affinity for the medium; they are typically inorganic or metallic.

Identify the key stabilizing factor

For lyophilic sols, the dominant stabilizing force is solvation — the solvent molecules form a protective layer around each particle. This layer is often several molecules thick and acts as a physical barrier, preventing particle-particle contact even if electrostatic repulsion is weak.

Evaluate each option

(A) “The colloidal particles are not solvated” — This is false. Lyophilic particles are solvated; that’s exactly why they’re stable. So (A) is incorrect.

(B) “There is a strong electrostatic interaction between the colloidal particles” — Electrostatic repulsion can help, but it’s not the main reason for lyophilic stability. In fact, lyophilic sols often have little or no charge; their stability comes from solvation. So (B) is not the best answer.

(C) “The colloidal particles have no charge” — Many lyophilic sols are uncharged (e.g., starch), but that doesn’t make them stable; it’s the solvation that matters. Also, some lyophilic sols do carry charge. So (C) is misleading and incorrect.

(D) “The Brownian movement” — Brownian motion keeps particles suspended but doesn’t explain relative stability between lyophilic and lyophobic sols. Both types exhibit Brownian motion. So (D) is irrelevant.

Conclude the correct reasoning …
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TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$Q. Lyophilic sols are more stable than lyophobic sols, because (A) The colloidal particles are not solvated (B) There is a strong electrostatic interaction between the colloidal particles (C) The colloidal particles have no change (D) The Brownian movement$
›Reveal solutionSolution
Lyophilic sols are more stable than lyophobic sols primarily because their particles are strongly solvated (coated with a solvent layer), which prevents coagulation — the correct option is (B) is not correct; the actual reason is solvation, so the answer is not among the given choices as stated, but the closest intended answer is (B) if interpreted as "strong interaction" meaning solvation, though the question's options are flawed. The standard answer is (B) in many textbooks, but we must explain carefully.

Watch out

Many students pick (A) because they misread "not solvated" — but lyophilic sols are solvated, so (A) is false. Option (C) is also false because lyophilic particles often carry charge. Option (D) is a general property of all colloids, not a distinguishing factor.

Concept and Intuition

The stability of a colloidal solution depends on how well the particles resist coming together (coagulating). Lyophilic ("solvent-loving") sols, like gelatin or starch in water, have particles that are strongly solvated — each particle is surrounded by a thick shell of solvent molecules. This solvation layer acts like a cushion, physically preventing particles from sticking even when they collide. In contrast, lyophobic ("solvent-hating") sols, like gold or silver in water, lack this protective layer; their stability relies mainly on electrostatic repulsion from surface charges, which is more easily disrupted by adding electrolytes.

Thus, the key reason lyophilic sols are more stable is solvation, not electrostatic repulsion or Brownian motion.

Step-by-Step Reasoning

Understand the terms

Lyophilic: particles have a strong affinity for the solvent (e.g., water-loving).

Lyophobic: particles have little or no affinity for the solvent. In water, lyophilic sols are called hydrophilic, lyophobic are hydrophobic.

Identify the main stabilizing factor for each

Lyophilic sols: stability comes from solvation — solvent molecules form a tightly bound layer around each particle. This layer is often many molecules thick and is thermodynamically favorable.

Lyophobic sols: stability comes from electrostatic repulsion — particles carry like charges, and the repulsive force prevents aggregation. This is more fragile; adding salt neutralizes charges and causes coagulation.

Evaluate each option

(A) "The colloidal particles are not solvated" — This is false for lyophilic sols; they are solvated. So (A) is incorrect. …
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Q.Why does copper not replace hydrogen from acids?

Concept understanding — Standard Reduction Potential

Cell Representation and the Nernst Equation: From Intuition to Precision

The Intuition First

The Precise Statement

Cell Representation: How We Write It

Applying the Nernst Equation to a Cell Representation

A Worked Example

Why this formula?

Standard Reduction Potential: Why the Formula Holds

1. The Core Idea: A Half-Cell's "Tendency to Gain Electrons"

2. The Formula: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

Why subtraction, not addition?

The reasoning step-by-step:

3. The Nernst Equation: Why $E = E^\circ - \frac{RT}{nF} \ln Q$

From Gibbs free energy:

Substituting:

Rearranging:

Why this makes physical sense:

The Concept: Standard Reduction Potential as a “Tug-of-War” for Electrons

Step-by-Step Reasoning

Method: Using the Electrochemical Series (Standard Reduction Potential Comparison)

Step 1: Recall the relevant half-reactions

Step 2: Identify the reaction direction

Step 3: Calculate the cell potential

Step 4: Interpret the result …

Common Mistakes: Why Copper Doesn't Replace Hydrogen from Acids

The Core Concept First

Mistake #1: Confusing Oxidation and Reduction Potentials

Mistake #2: Memorizing Without Understanding the Series

Mistake #3: Forgetting the Role of Oxidizing Acids

Concept and Intuition

Step-by-Step Reasoning

Q.Why does copper not replace hydrogen from acids?

The Concept: Standard Reduction Potential as a “Tug-of-War” for Electrons

Step-by-Step Reasoning

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