Two metals \(A\) and \(B\) have \(E_{\mathrm{RP}}^{\circ}=+0.76 \mathrm{~V}\) and \(-0.80 \mathrm{~V}\) respectively. Which will liberate \(\mathrm{H}_{2}\) from \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Short Answer

Expert verified
Metal A with a standard reduction potential of +0.76 V will liberate hydrogen gas from \(H_2SO_4\), as its potential is higher than both Metal B and hydrogen ions in the acid.

Step by step solution

01

Understanding Standard Reduction Potentials

The Standard Reduction Potentials (\(E_{\text{RP}}^\text{o}\)) indicate the willingness of a metal to be reduced, meaning to gain electrons. A higher (more positive) reduction potential indicates a greater tendency to gain electrons, while a more negative potential indicates a lesser tendency.
02

Comparing the Reduction Potentials of Metals A and B

Metal A has a reduction potential of +0.76 V, which is more positive compared to metal B, which has a reduction potential of -0.80 V. This means Metal A has a higher tendency to be reduced than metal B.
03

Determining the Ability to Liberate Hydrogen Gas

Since liberating hydrogen gas from an acid (like \(H_2SO_4\)) involves a reduction process (2H+ + 2e- \rightarrow H2), the metal that will actually liberate \(H_2\) must have a more positive reduction potential than the \(H^+\) ions in the acid. The reduction potential of \(H^+\) to \(H_2\) is 0 V. Therefore, Metal A with +0.76 V can liberate \(H_2\), as it has a higher (more positive) reduction potential than 0 V.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Electrochemistry
Electrochemistry is a branch of chemistry that studies the relationship between electricity and chemical reactions. This field is crucial for understanding how batteries work, the principles behind corrosion, and the functioning of various sensors. At its core, electrochemistry is concerned with the movement of electrons between molecules in redox reactions, which are fundamental to generating electrical energy from chemical processes, and vice versa.

One of the fundamental concepts within electrochemistry is the standard reduction potential (\(E_{\text{RP}}^\text{o}\)), which measures the tendency of a chemical species to acquire electrons and thus be reduced. The standard reduction potentials are tabulated under standard conditions, which include a temperature of 298 K, a 1 M concentration for each ion participating in the reaction, a partial pressure of 1 atm for any gases that are involved, and a pH of 7 or neutral. These potentials are measured in volts (V) and are relative to the standard hydrogen electrode (SHE) which has a potential of 0 V.
The Role of Redox Reactions
Redox reactions are a type of chemical reaction involving the transfer of electrons between two species. The term 'redox' comes from two concepts - reduction and oxidation. Reduction refers to the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion; oxidation is the loss of electrons or an increase in oxidation state. These reactions are essential because they are the basis for the production of energy in batteries and are also involved in many biological processes.

For instance, the reaction of a metal with sulfuric acid to liberate hydrogen gas, as mentioned in the exercise, is a typical redox reaction where the sulfuric acid (H2SO4) provides protons (H+) that are reduced, meaning they gain electrons to form hydrogen gas (\(H_2\)). This type of reaction occurs often in electrochemical cells where the oxidation state of elements changes. The metal that is more easily oxidized (loses electrons) during these redox processes plays a critical role in determining which direction the reaction proceeds.
Chemical Properties of Metals and Their Reactions
The chemical properties of metals can vary greatly, but they often share some common traits such as good electrical and thermal conductivity, luster, and malleability. In the context of electrochemistry, one of the key properties to consider is the reduction potential of a metal. As we've seen in the exercise, the standard reduction potentials determine how easily a metal can be reduced or, in other words, gain electrons.

Metal A, with a higher reduction potential than metal B, indicates that it is more willing to accept electrons than Metal B or the protons from an acid like sulfuric acid. The ability of a metal to lose or gain electrons influences its reactivity with other substances, including whether or not it can displace hydrogen from an acid to produce hydrogen gas. This concept highlights why some metals, like calcium or magnesium, react vigorously with acids liberating hydrogen, while others like gold or platinum, do not react so readily, if at all.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

$$ \text { Write equations for the electrolysis of } \mathrm{CaH}_{2} \text { in fused state. } $$

A metal is known to form fluoride \(M \mathrm{~F}_{2}\). When 10 ampere electricity is passed throwgh i multen salt for \(330 \mathrm{sec}, 1.95 \mathrm{~g}\) metal is deposited. Find out the atomic weight of metal. What will be the quantity of charge required to deposit the same mass of \(\mathrm{Cu}\) from \(\mathrm{CuSO}_{4}\) (aq.) (At. wt. of \(\mathrm{Cu}=63.6\) )

Given the standard electrode potentials; \(\mathrm{K}^{+} / \mathrm{K}=-2.93 \mathrm{~V}, \quad \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}, \quad \mathrm{Hg}^{2+} / \mathrm{Hg}=0.79 \mathrm{~V}\) \(\mathrm{Mg}^{2+} / \mathrm{Mg}=-2.37 \mathrm{~V}, \quad \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.74 \mathrm{~V}\) Arrange these metals in their increasing order of reducing power.

$$ \begin{aligned} &\text { The e.m.f. of the cell } M \mid M^{n+}(0.02 M) \| \mathrm{H}^{+}(1 M) \mathrm{H}_{2(\mathrm{~g})}(1 \mathrm{~atm}) \mathrm{Pt} \text { at }\\\ &25^{\circ} \mathrm{C} \text { is } 0.81 \mathrm{~V} \text { . Calculate the valence of metal if } E^{\circ}{ }_{\ldots 1 / 2}^{n+}=0.76 \mathrm{~V} \end{aligned} $$

For the cell : $$ \text { Zn }\left|\begin{array}{c} \mathrm{Zn}_{\text {aq }}^{2+} \\ 1 M \end{array}\right|\left|\begin{array}{c|c} \mathrm{Cu}_{a q}^{2+} \\ 2 M \end{array}\right| \mathrm{Cu} $$ Calculate the values for ; (a) cell reaction, (b) \(E_{\text {cell }}^{\circ}\). (c) \(E_{\text {cell }}\) (d) the minimum concentration of \(\mathrm{Cu}^{2+}\) at which cell reaction is spontaneous if \(\mathrm{Zn}^{2+}\) is \(1 M\) (e) does the displacement of \(\mathrm{Cu}^{2+}\) goes almost to completion. Given : \(E_{R P_{\mathrm{Cu}^{2+} / \mathrm{Cu}}}^{\circ}=+0.35 \mathrm{~V}\) $$ E_{R P_{\mathrm{Zn}^{2+} / Z n}}^{\circ}=-0.76 \mathrm{~V} $$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free