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Lostsunrise [7]
3 years ago
6

What is the standard reduction potential,E, for the half-reaction Zn2+(aq)+2e- -> Zn(s)?

Chemistry
1 answer:
Tatiana [17]3 years ago
8 0

Answer:

E = -0.76 V

Explanation:

The potential between the anode and the cathode is called standard reduction potential. It gives the tendency of the species to get reduced.

The given half reaction is :

Zn^{2+}(aq)+2e^{-}\rightarrow Zn(s)

Zn²⁺ is the reduced form of Zn.

It is a stronger reducng agent.

Its standard reduction potential (E) has a sandard value.

E = -0.76 V

Hence, the correct option is (c).

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Write the net ionic equation for the reaction of aqueous solutions of ammonium chloride and iron(III) hydroxide.
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Answer:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

Explanation:

Hello.

In this case, for the reaction between aqueous solutions of ammonium chloride and iron (III) hydroxide, we have the following complete molecular reaction:

3NH_4Cl(aq)+Fe(OH)_3(s)\rightarrow 3NH_4OH+FeCl_3

And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:

3NH_4^+(aq)+3Cl^-(aq)+Fe(OH)_3(s)\rightarrow 3NH_4^+(aq)+3OH^-(aq)+Fe^{3+}(aq)+3Cl^-(aq)

Finally, the net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ as they are both the left and right side, therefore, the net ionic equation is:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

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Answer:

2 mol H

Explanation:

For every 2 mol of NaOH, we're reacting 2 mol of H2O. In order to figure out how many mol of H are needed, it needs to be set up stochiometrically. Starting off with the given value, 1 mol of NaOH, we can then make a mol to mol ratio. For 2 mol of NaOH, we have 2 mol of H2O. For every 2 mol of H2O, we have 4 mol of H (this is because we are multiplying the coefficient by the subscript: 2 × 2). Now, we can solve for our answer.

1 mol NaOH × (2 mol H₂O / 2 mol NaOH) × (4 mol H / 2 mol H₂O)

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