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Chapter 18: Problem 80

An electrolytic cell consists of a \(100.0-\mathrm{g}\) strip of copper in \(0.200 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) and a \(100.0-\mathrm{g}\) strip of \(\mathrm{Cr}\) in \(0.200 \mathrm{M} \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\). The overall reac- tion is: $$ 3 \mathrm{Cu}(s)+2 \mathrm{Cr}^{3+}(a q) \longrightarrow 3 \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cr}(s) \quad E^{\circ}=-1.083 \mathrm{~V} $$ An external battery provides 3 amperes for 70 minutes and 20 seconds with \(100 \%\) efficiency. What is the mass of the copper strip after the battery has been disconnected?

Short Answer

Expert verified
Answer: The mass of the copper strip after the battery has been disconnected is 87.50 g.

Step by step solution

01

Convert Time to Seconds

Firstly, we need to convert the time the battery was active (70 minutes 20 seconds) into seconds for further calculations: $$ 70~\text{minutes}~20~\text{seconds} = 70 \times 60 + 20 = 4220~\text{seconds} $$
02

Calculate the Charge Passed

We are given the current (3 amperes), and we have the time in seconds now. We can use the following formula to find the total charge (Q) passed through the circuit: $$ Q = I \times t $$ where Q is the total charge, I is the current, and t is the time. Plugging in the values, we get: $$ Q = 3~\text{A} \times 4220~\text{seconds} = 12660~\text{Coulombs} $$
03

Calculate Moles of Electrons Transferred

According to Faraday's Law, the moles of electrons (n) transferred are given by: $$ n = \frac{Q}{F} $$ where F is the Faraday constant, approximately equal to \(96500 ~\text{C/mol}\). Plug in the values and calculate: $$ n = \frac{12660~\text{C}}{96500~\text{C/mol}} = 0.13118~\text{mol} $$
04

Calculate Moles of Copper Lost

Now, we will use the stoichiometry of the equation to determine the moles of copper (Cu) lost. The balanced equation shows that 3 moles of Cu are lost for each 2 moles of electrons transferred: $$ \text{Moles of Cu lost} = \frac{3 \text{moles of Cu}}{2 \text{moles of e-}} \times 0.13118~\text{mol} = 0.19677~\text{mol} $$
05

Calculate the Mass of Copper Lost

We will now convert the moles of Cu lost to mass using the molecular weight (63.5 g/mol): $$ \text{Mass of Cu lost} = 0.19677~\text{mol} \times 63.5~\text{g/mol} = 12.50~\text{g} $$
06

Calculate the Mass of the Copper Strip after Electrolysis

Finally, subtract the mass of Cu lost from the initial mass of the Cu strip: $$ \text{Mass of Cu strip after electrolysis} = 100.0~\text{g} - 12.50~\text{g} = 87.50~\text{g} $$ The mass of the copper strip after the battery has been disconnected is 87.50 g.

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Most popular questions from this chapter

Which species in each pair is the stronger reducing agent? (a) \(\mathrm{Cr}\) or \(\mathrm{Cd}\) (b) \(\mathrm{I}^{-}\) or \(\mathrm{Br}^{-}\) (c) \(\mathrm{OH}^{-}\) or \(\mathrm{NO}_{2}^{-}\) (d) NO in acidic solution or NO in basic solution

I Suppose \(E_{\text {red }}^{\circ}\) for \(\mathrm{H}^{+} \longrightarrow \mathrm{H}_{2}\) were taken to be \(0.300 \mathrm{~V}\) instead of \(0.000 \mathrm{~V}\). What would be (a) \(E_{\text {ox }}^{o}\) for \(\mathrm{H}_{2} \longrightarrow \mathrm{H}^{+}\) ? (b) \(E_{\text {red }}^{\circ}\) for \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-} ?\) (c) \(E^{\circ}\) for the cell in \(24(\mathrm{c})\) ? Compare your answer with that obtained in \(24(\mathrm{c})\).

Consider the following reaction carried out at \(1000^{\circ} \mathrm{C}\). $$ \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) $$ Assuming that all gases are at \(1.00 \mathrm{~atm}\), calculate the voltage produced at the given conditions. (Use Appendix 1 and assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with an increase in temperature.)

Consider a salt bridge voltaic cell represented by the following reaction: $$ \mathrm{Ca}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2}(g) $$ (a) What is the direction of the electrons in the external circuit? (b) What electrode can be used at the cathode? (c) What is the reaction occurring at the anode?

A baby's spoon with an area of \(6.25 \mathrm{~cm}^{2}\) is plated with silver from \(\mathrm{AgNO}_{3}\) using a current of \(2.00 \mathrm{~A}\) for two hours and 25 minutes. (a) If the current efficiency is \(82.0 \%\), how many grams of silver are plated? (b) What is the thickness of the silver plate formed \(\left(d=10.5 \mathrm{~g} / \mathrm{cm}^{3}\right)\) ?
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