Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An unknown metal M is electrolyzed. It took 74.1 s for a current of 2.00 A to plate out 0.107 g of the metalfrom a solution containing M(NO3)3. Identify the metal.

Short Answer

Expert verified

Theunknown metal is bismuth

Step by step solution

01

Given information

The mass of plate-out metal is 0.107 g. The time value is 74.1 s. The current can be given as 2 A. The reaction for Ru formation can be given as:M3++3e-M

02

Mole number of electrons released

The number of moles of electron released with 2.5 A current and 50 mins can be expressed as:

Molesofelectron=74.1s2C1s1mol96485C=0.00154mol

The moles of Ruthenium produced in electrolysis can be expressed as follows:

Mole=0.00154mol1mol3mol=5.13×10-4

03

Finding molar mass

The molar mass can be given as:

Molarmass=Givenmassmoles=0.107g5.13×10-4molMolarmass=208.6g/mol

The metal with the above molar mass is bismuth.

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!

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

The amount of manganese in steel is determined bychanging it to permanganate ion. The steel is first dissolved in nitric acid, producing Mn²+ ions. These ionsare then oxidized to the deeply colored MnO-4 ions by
periodate ionIO4- in acid solution.
a. Complete and balance an equation describing eachof the above reactions.
b. Calculate Eo,ΔGo and K at 25°C for each reaction.

Sketch the galvanic cells based on the following half-reactions. Calculate, show the direction of electron flow and the direction of ion migration through the salt bridge, identify the cathode and anode, and give the overall balanced equation. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm.

a.CI2+2e-2CI-,E0=1.36VBr2+2e-2Br-,E0=1.09Vb.MnO4+8H++5e-Mn2++4H2O,E0=1.51VIO4-+2H++2e-IO3-+H2O,E0=1.60Vc.H2O2+2H++2e-2H2O,E0=1.78VO2+2H++2e-H2O2,E0=0.68Vd.Mn2++2e-Mn,E0=-1.18VFe3++3e-Fe,E0=-0.036V

It takes 15 kWh (kilowatt hours) of electrical energy toproduce 1.0 kg of aluminum metal from aluminium oxide by the Hall-Heroult process. Compare this value with the amount of energy necessary to melt 1.0 kg of aluminum metal. Why is it economically feasible to recycle aluminum cans? (The enthalpy of fusion for aluminium metal is 10.7 kJ/mol and 1 watt = 1 J/s.)

An electrochemical cell consists of a nickel metal electrode immersed in a solution with [Ni2+] = 1.0 M separated by a porous disk from an aluminum metal electrode immersed in a solution with [Al3+] = 1.0 M. Sodium hydroxide is added to the aluminum compartment, causing Al(OH)3(s) to precipitate. After precipitation of Al(OH)3has ceased, the concentration of OH is 1.0 x 10-4 M, and the measured cell potential is 1.82 V. Calculate the Ksp value for Al(OH)3.

Al(OH)3(s)Al3+(aq)+3OH-(aq)Ksp=?

Question:A galvanic cell is based on the following half-reactions:

Fe+2+2e-Fe(s)E=-0.4402H++2e-H2(g)E=0

In this cell, the iron compartment contains an iron electrodeand [Fe+2]=1.00×10-3atm, and the hydrogen compartment contains a platinum electrode,role="math" localid="1663906376319" pH2=1.00atmand a weak acid HA at an initial concentration of role="math" localid="1663906339580" 1.00M.If the observed cell potential is 0.333Vat 25C, calculate the Ka value for the weak acid HA atrole="math" localid="1663906523400" 25C.

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