115 Intermolecular Force Worksheet Key - Intermolecular Force ... - Free Printable

Let’s solve this step by step.

We are given three half-reactions with their standard reduction potentials (E°):

1. Fe³⁺(aq) + e⁻ → Fe²⁺(aq) E° = +0.77 V
2. Cu²⁺(aq) + 2e⁻ → Cu(s)   E° = +0.34 V
3. Zn²⁺(aq) + 2e⁻ → Zn(s)   E° = -0.76 V

We need to find which pair of these can be combined to make a galvanic cell that produces exactly 1.10 V.

In a galvanic cell, the overall voltage is:

E°_cell = E°_cathode – E°_anode

The cathode is where reduction happens (higher E°), and the anode is where oxidation happens (lower E°). So we subtract the smaller E° from the larger one.

Let’s try all possible pairs:

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Pair 1: Fe³⁺/Fe²⁺ and Cu²⁺/Cu

Cathode: Fe³⁺/Fe²⁺ (+0.77 V)
Anode: Cu²⁺/Cu (+0.34 V) — but wait! If we use Cu as anode, it would be oxidized: Cu → Cu²⁺ + 2e⁻, so its potential for oxidation is -0.34 V? No — better to think:

E°_cell = E°_reduction(cathode) – E°_reduction(anode)

So if Fe³⁺ is reduced (cathode), and Cu is oxidized (so Cu²⁺/Cu is the anode half-cell), then:

E°_cell = 0.77 – 0.34 = +0.43 V → Not 1.10 V

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Pair 2: Fe³⁺/Fe²⁺ and Zn²⁺/Zn

Cathode: Fe³⁺/Fe²⁺ (+0.77 V)
Anode: Zn²⁺/Zn (-0.76 V)

E°_cell = 0.77 – (-0.76) = 0.77 + 0.76 = +1.53 V → Too high

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Pair 3: Cu²⁺/Cu and Zn²⁺/Zn

Cathode: Cu²⁺/Cu (+0.34 V)
Anode: Zn²⁺/Zn (-0.76 V)

E°_cell = 0.34 – (-0.76) = 0.34 + 0.76 = +1.10 V ✔ Perfect!

This matches exactly what we want.

Now, let’s write the balanced overall reaction.

At cathode (reduction):
Cu²⁺(aq) + 2e⁻ → Cu(s)

At anode (oxidation):
Zn(s) → Zn²⁺(aq) + 2e⁻

Add them together (electrons cancel):

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

That’s the net ionic equation.

Now, check the options in part (b) — they list several combinations. We’re looking for the one that gives 1.10 V and has the correct reactants/products.

Looking at option D:

> D. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
> The voltage produced by this reaction is 1.10 V.

Yes! That’s exactly what we got.

Also, note: In the question, it says “which combination... will produce a cell with a voltage of 1.10 V?” and then lists A–F. Only D matches our calculation.

Wait — let me double-check if any other pair could give 1.10 V.

What if we reversed Fe and Zn? We already did — that gave 1.53 V.

What if we used Fe as anode and Cu as cathode? That was 0.43 V.

No other combination gives 1.10 V.

Also, in part (a), it asks to identify strongest/weakest reducing agents.

Strongest reducing agent = most easily oxidized = lowest (most negative) E° reduction → Zn(s) (since Zn²⁺/Zn is -0.76 V, meaning Zn metal wants to lose electrons badly)

Weakest reducing agent = hardest to oxidize = highest E° reduction → Fe²⁺? Wait no — the species listed are metals or ions?

Actually, the reducing agents are the ones that get oxidized — so for each half-reaction, the left side (reactant) when written as reduction is the oxidizing agent; the right side (product) is the reducing agent.

But typically, we say:

- Strongest reducing agent: the metal with most negative E° → Zn(s)
- Weakest reducing agent: the ion that is hardest to reduce? Or the metal that doesn’t want to oxidize?

Standard way:

Reducing strength increases as E° reduction decreases.

So:

Strongest reducing agent: Zn(s) (E° = -0.76 V for Zn²⁺/Zn)

Then Cu(s) (E° = +0.34 V for Cu²⁺/Cu)

Then Fe²⁺? Wait — Fe²⁺ is not a reducing agent here; Fe³⁺/Fe²⁺ couple — Fe²⁺ can act as reducing agent to become Fe³⁺.

Actually, for the couples given:

- For Fe³⁺ + e⁻ → Fe²⁺, the reducing agent is Fe²⁺ (it can be oxidized to Fe³⁺)
- For Cu²⁺ + 2e⁻ → Cu, reducing agent is Cu(s)
- For Zn²⁺ + 2e⁻ → Zn, reducing agent is Zn(s)

So comparing reducing strengths:

Zn(s) > Cu(s) > Fe²⁺ ? Let’s see their tendencies to be oxidized.

E° for oxidation:

Zn → Zn²⁺ + 2e⁻ : E°_ox = +0.76 V
Cu → Cu²⁺ + 2e⁻ : E°_ox = -0.34 V
Fe²⁺ → Fe³⁺ + e⁻ : E°_ox = -0.77 V

Higher E°_ox means stronger reducing agent.

So:

Zn(s): +0.76 V → strongest
Cu(s): -0.34 V
Fe²⁺: -0.77 V → weakest

So strongest reducing agent: Zn(s)
Weakest reducing agent: Fe²⁺

But in the worksheet, part (a) says:

"Identify the strongest reducing agent from the product or reactant sides..."

And lists:

Reactant side: Fe³⁺, Cu²⁺, Zn²⁺ — these are oxidizing agents, not reducing.

Product side: Fe²⁺, Cu(s), Zn(s) — these can be reducing agents.

So among products: Fe²⁺, Cu(s), Zn(s)

As above, Zn(s) is strongest reducing agent, Fe²⁺ is weakest.

Similarly, strongest oxidizing agent: highest E° reduction → Fe³⁺ (+0.77 V)

Weakest oxidizing agent: lowest E° reduction → Zn²⁺ (-0.76 V)

But the question only asks for strongest and weakest reducing agents in part (a).

Anyway, back to part (b) — we’ve confirmed that D is correct.

Now, just to be thorough, let’s look at why others are wrong.

A: Uses Fe³⁺ and Cu — voltage 0.43 V, not 1.10

B: Fe³⁺ and Zn — 1.53 V

C: Says "strongest oxidizing agent is Fe³⁺" — true, but then says voltage is 1.53 V for Fe³⁺ and Zn — which is correct for that pair, but not 1.10 V

D: Correct as we saw

E: Says hydrogen reading — irrelevant, and voltage 0.77 V? Not matching

F: Says battery uses copper and zinc — yes, but then says voltage is 1.10 V — actually, that’s correct, but the reaction written is Zn + Cu²⁺ → Zn²⁺ + Cu, same as D. But in F, it says "the voltage produced by this reaction is 1.10 V" — which is true, but let’s read carefully.

In F: "A battery that uses copper and zinc electrodes... The voltage produced by this reaction is 1.10 V."

It doesn’t specify the reaction, but implies the standard Daniell cell, which is indeed 1.10 V. However, in the context, D explicitly writes the reaction and says 1.10 V, while F might be ambiguous.

But looking back at the image text (though I shouldn't describe it, but based on common knowledge), option D is precisely the reaction we derived.

Moreover, in multiple choice, usually only one is fully correct.

But let me check the exact wording in the user's input.

In the user's message, for part (b), option D says:

"D. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
The voltage produced by this reaction is 1.10 V."

Which is correct.

Option F says:

"F. A battery that uses copper and zinc electrodes... The voltage produced by this reaction is 1.10 V."

But it doesn't specify the reaction, and also says "hydrogen gas is not involved", which is true, but it's less specific.

However, since D directly states the reaction and voltage correctly, and matches our calculation, D is the best answer.

Additionally, in some contexts, F might be considered correct, but given that D is explicit and accurate, and the question likely expects D.

But wait — in the initial problem, it says "which combination of the following reactions" — and lists A-F as descriptions.

Upon second thought, both D and F refer to the same cell, but D gives the chemical equation, while F describes it verbally.

But in D, it says "the voltage produced by this reaction is 1.10 V" — which is correct.

In F, it says "the voltage produced by this reaction is 1.10 V" — also correct, but "this reaction" refers to the battery using copper and zinc, which is the same.

However, looking at the options, probably only D is listed with the correct equation and voltage.

To resolve, let's calculate again: E°_cell for Zn/Cu is indeed 0.34 - (-0.76) = 1.10 V.

And the reaction is Zn + Cu²⁺ → Zn²⁺ + Cu.

So D is correct.

Now, for part (a), although not asked in the final answer selection, but since the user included it, but the main task is part (b).

The user's instruction is to solve the problem, and the image has parts a and b, but the critical part is identifying which option in b is correct.

Given that, and since the question says "solve the problem accurately", and the problem includes selecting the correct option in b.

Moreover, in the response format, we need to provide final answer.

So, after careful reasoning, the correct choice is D.

Final Answer: D