Chapter 19: Problem 43
What is the difference between a galvanic cell (such as a Daniell cell) and an electrolytic cell?
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 19: Problem 43
What is the difference between a galvanic cell (such as a Daniell cell) and an electrolytic cell?
All the tools & learning materials you need for study success - in one app.
Get started for free
Gold will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. However, the metal does dissolve in a mixture of the acids (one part \(\mathrm{HNO}_{3}\) and three parts \(\mathrm{HCl}\) by volume ), called aqua regia. (a) Write a balanced equation for this reaction. (Hint: Among the products are \(\mathrm{HAuCl}_{4}\) and \(\mathrm{NO}_{2} .\) ) (b) What is the function of \(\mathrm{HCl} ?\)
What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M}\) \(\left[\mathrm{H}^{+}\right]=0.050 \mathrm{M},\) and \(\mathrm{P}_{\mathrm{H}_{2}}=1.0 \mathrm{~atm} ?\)
Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.
A constant electric current flows for \(3.75 \mathrm{~h}\) through two electrolytic cells connected in series. One contains a solution of \(\mathrm{AgNO}_{3}\) and the second a solution of \(\mathrm{CuCl}_{2}\). During this time \(2.00 \mathrm{~g}\) of silver are deposited in the first cell. (a) How many grams of copper are deposited in the second cell? (b) What is the current flowing, in amperes?
A sample of iron ore weighing \(0.2792 \mathrm{~g}\) was dissolved in an excess of a dilute acid solution. All the iron was first converted to \(\mathrm{Fe}(\mathrm{II})\) ions. The solution then required \(23.30 \mathrm{~mL}\) of \(0.0194 \mathrm{M} \mathrm{KMnO}_{4}\) for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.
What do you think about this solution?
We value your feedback to improve our textbook solutions.