Solution Found!
Calculate the voltage of each of the following cells. With the reasoning in Figure 13-8
Chapter 13, Problem 13-C(choose chapter or problem)
Calculate the voltage of each of the following cells. With the reasoning in Figure 13-8, state the direction of electron flow.
(a) \(\mathrm{Fe}(s)\left|\mathrm{FeBr}_{2}(0.010 \mathrm{M}) \| \mathrm{NaBr}(0.050 \mathrm{M})\right| \mathrm{Br}_{2}(l) \mid \operatorname{Pt}(s)\)
(b) \(\mathrm{Cu}(s)\left|\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(0.020 \mathrm{M}) \| \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(0.050 \mathrm{M})\right| \mathrm{Fe}(s)\)
(c) \(\mathrm{Hg}(l)\left|\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\right| \mathrm{KCl}(0.060 \mathrm{M}) \| \mathrm{KCl}(0.040 \mathrm{M}) \mid \mathrm{Cl}_{2}(g, 0.50 \text { bar }) \mid \operatorname{Pt}(s)\)
Questions & Answers
QUESTION:
Calculate the voltage of each of the following cells. With the reasoning in Figure 13-8, state the direction of electron flow.
(a) \(\mathrm{Fe}(s)\left|\mathrm{FeBr}_{2}(0.010 \mathrm{M}) \| \mathrm{NaBr}(0.050 \mathrm{M})\right| \mathrm{Br}_{2}(l) \mid \operatorname{Pt}(s)\)
(b) \(\mathrm{Cu}(s)\left|\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(0.020 \mathrm{M}) \| \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(0.050 \mathrm{M})\right| \mathrm{Fe}(s)\)
(c) \(\mathrm{Hg}(l)\left|\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\right| \mathrm{KCl}(0.060 \mathrm{M}) \| \mathrm{KCl}(0.040 \mathrm{M}) \mid \mathrm{Cl}_{2}(g, 0.50 \text { bar }) \mid \operatorname{Pt}(s)\)
ANSWER:Step 1 of 4
In an electrochemical cell, the anode of the cell undergoes oxidation and releases electrons which will be collected by cathode and undergoes reduction. In an electrochemical cell, the conventional placing of the anode is on the left side.Thus, electrons will flow from the left to the right. This is the reason why the flow of electrons is to the right in the figure 13-8 given.