The following concentration cell is constructed. \(\mathrm{Ag} | \mathrm{Ag}^{+}\left(0.10 \mathrm{M}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}, 0.10 \mathrm{M} \mathrm{CN}^{-}\right)\) $$\| \mathrm{Ag}^{+}(0.10 \mathrm{M}) | \mathrm{Ag}$$ If \(K_{f}\) for \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) is \(5.6 \times 10^{18},\) what value would you expect for \(E_{\text {cell }}\) ? [Hint: Recall that the anode is on the left.]

Short Answer

Expert verified
The expected value for \(E_{\text {cell }}\) is -0.0296 V

Step by step solution

01

Identify the half-reactions

Let's identify the half-reactions. For the left half-cell, \(\mathrm{Ag}(\mathrm{CN})_{2}\) is converted into Ag and this half-reaction can be represented as: \[\mathrm{Ag}(\mathrm{CN})_{2}^{-} + e^{-} \rightarrow \mathrm{Ag} + 2 \mathrm{CN}^{-}\] Similarly for the right half cell, we have: \[\mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag}\]
02

Calculate the half-cell reduction potentials

Let’s understand that the left half-cell is the anode where oxidation occurs and the right half-cell is the cathode where reduction occurs. According to the Nernst equation, the electromotive force (EMF) of a half-cell reaction at non-standard conditions can be represented as: \[E = E^{\circ} - \frac{0.0592}{n} \log Q\] where \(E^{\circ}\) is the standard reduction potential, n is the number of electron transfer, and Q is the reaction quotient. For the left half-cell, we write the reaction quotient (Q) as \([Ag][CN^-]^2/ [Ag(CN)_2^-] = 0.0010\), given that the concentration of Ag, which is a solid, is counted as 1 in the equilibrium expression. For the right half-cell, as the reaction is under standard conditions, \(Q = 1\) and hence the Ecell is simply the standard reduction potential. Note that the value of \(E^{\circ}\) for both half reactions is given as zero since Ag+ ions are involved in both half-cells.
03

Plug the values into the Nernst equation

Now, we can plug values into the Nernst equation to get E for both half-cells. For left half cell, we have \(E_{\text{left}} = 0 - \frac{0.0592}{1} \log (0.0010) = 0.0296 V\). For the right half cell under standard conditions, we have \(E_{\text{right}} = 0 V\). The overall cell potential can be calculated as the difference between the reduction potentials at the cathode and anode, i.e. \(E_{\text{cell}} = E_{\text{right}} - E_{\text{left}} = 0 - 0.0296 = -0.0296 V\).
04

Convert the final answer to required unit

The final electromotive potential should be determined in Volt (V). Note that the negative sign in the answer indicates the cell is not spontaneous in the given direction. In order to get a positive cell potential, the cell would need to be reversed.

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

Write the formula and name of (a) a complex ion having \(\mathrm{Cr}^{3+}\) as the central ion and two \(\mathrm{NH}_{3}\) molecules and four \(\mathrm{Cl}^{-}\) ions as ligands (b) a complex ion of iron(III) having a coordination number of 6 and \(\mathrm{CN}^{-}\) as ligands (c) a coordination compound comprising two types of complex ions: one a complex of \(\mathrm{Cr}(\mathrm{III})\) with ethylenediamine (en), having a coordination number of 6 the other, a complex of \(\mathrm{Ni}(\mathrm{II})\) with \(\mathrm{CN}^{-}\), having a coordination number of 4

In both \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) ions, the iron is present as \(\mathrm{Fe}(\mathrm{II}) ;\) however, \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is paramagnetic, whereas \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) is diamagnetic. Explain this difference.

A structure that Werner examined as a possible alternative to the octahedron is the trigonal prism. (a) Does this structure predict the correct number of isomers for the complex ion \(\left[\mathrm{CoCl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\right]^{+} ?\) If not, why not? (b) Does this structure account for optical isomerism in \(\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+} ?\) Explain.

Briefly describe each of the following ideas, phenomena, or methods: (a) spectrochemical series; (b) crystal field theory; (c) optical isomer; (d) structural isomerism.

Draw Lewis structures for the following ligands: (a) \(\mathrm{H}_{2} \mathrm{O} ;\) (b) \(\mathrm{CH}_{3} \mathrm{NH}_{2} ;\) (c) \(\mathrm{ONO}^{-} ;\) (d) SCN \(^{-}\).

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