/*! 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} Problem 61 A voltaic cell is constructed th... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 61

A voltaic cell is constructed that uses the following reaction and operates at \(298 \mathrm{~K}\) : $$ \mathrm{Zn}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Ni}(s) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ni}^{2+}\right]=3.00 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ni}^{2+}\right]=0.200 M\), and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)

Short Answer

Expert verified
(a) Under standard conditions, the emf of the cell is 0.50 V. (b) With [\(\text{Ni}^{2+}\)] = 3.00 M and [\(\text{Zn}^{2+}\)] = 0.100 M, the emf of the cell is approximately 0.63 V. (c) With [\(\text{Ni}^{2+}\)] = 0.200 M and [\(\text{Zn}^{2+}\)] = 0.900 M, the emf of the cell is approximately 0.44 V.

Step by step solution

01

For the given reaction: \[ \text{Zn}(s) + \text{Ni}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Ni}(s) \] We can split it into two half-reactions: 1. Oxidation of Zn: \[ \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \] 2. Reduction of Ni虏鈦�: \[ \text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \] #Step 2: Find the standard reduction potentials#

From a standard reduction potentials table, we can find the values for both half-reactions: 1. For Zn: \[ E_{\text{Zn}^2+/\text{Zn}}^0 = -0.76 \,\text{V} \] 2. For Ni虏鈦�: \[ E_{\text{Ni}^2+/\text{Ni}}^0 = -0.26 \,\text{V} \] #Step 3: Calculate the standard emf of the cell#
02

The standard emf of the cell is the difference between the standard reduction potentials of the two half-reactions: \[ E_{\text{cell}}^0 = E_{\text{Ni}^2+/\text{Ni}}^0 - E_{\text{Zn}^2+/\text{Zn}}^0 \]

(a) Calculate the emf under standard conditions#
03

Plugging in the standard reduction potentials we found in step 2: \[ E_{\text{cell}}^0 = (-0.26\,\text{V}) - (-0.76\,\text{V}) = 0.50\,\text{V} \] Under standard conditions, the emf of the cell is 0.50 V. #Step 4: Use the Nernst equation#

We will now use the Nernst equation to calculate the emf of the cell with the given concentrations: \[ E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0592\,\text{V}}{n} \log_{10}\left(\frac{[\text{Zn}^{2+}]}{[\text{Ni}^{2+}]}\right) \] Where \(n\) is the number of electrons transferred in the reaction, which is 2 in this case.
04

(b) Calculate the emf with given concentrations#

Plug in the given concentrations and the calculated standard emf into the Nernst equation: \[ E_{\text{cell}} = 0.50\,\text{V} - \frac{0.0592\,\text{V}}{2} \log_{10}\left(\frac{0.100\,\text{M}}{3.00\,\text{M}}\right) \] \[ E_{\text{cell}} \approx 0.63\,\text{V} \] With [\(\text{Ni}^{2+}\)] = 3.00 M and [\(\text{Zn}^{2+}\)] = 0.100 M, the emf of the cell is approximately 0.63 V.
05

(c) Calculate the emf with different concentrations#

Plug in the given concentrations and the calculated standard emf into the Nernst equation: \[ E_{\text{cell}} = 0.50\,\text{V} - \frac{0.0592\,\text{V}}{2} \log_{10}\left(\frac{0.900\,\text{M}}{0.200\,\text{M}}\right) \] \[ E_{\text{cell}} \approx 0.44\,\text{V} \] With [\(\text{Ni}^{2+}\)] = 0.200 M and [\(\text{Zn}^{2+}\)] = 0.900 M, the emf of the cell is approximately 0.44 V.

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