Question

For each of the following electrochemical cells, calculate the Eo, write the balanced cell reaction equation and identify what is being oxidized and reduced, and calculate the Q and Ecell. Note that the substances with M given are in solution. All other substances are solids.

a. Cd (s)| Cd(NO3)2 (1.24 M) || FeCl2 (0.500 M), FeCl3 (0.750 M) | Pt (s)

b. Ag, Ag2CrO4 (s)| Na2CrO4 (2.00 M) || Mn(NO3)2 (0.100 M), HNO3 (2.00 M) | MnO2, Pt(s)

Answer 1

Answer on Question #56849 - Chemistry - General Chemistry

Question:

For each of the following electrochemical cells, calculate the Eo, write the balanced cell reaction equation and identify what is being oxidized and reduced, and calculate the Q and Ecell. Note that the substances with M given are in solution. All other substances are solids.

a. Cd (s) | Cd(NO3)2 (1.24 M) | FeCl2 (0.500 M), FeCl3 (0.750 M) | Pt (s)

b. Ag, Ag2CrO4 (s) | Na2CrO4 (2.00 M) | Mn(NO3)2 (0.100 M), HNO3 (2.00 M) | MnO2, Pt(s)

Answer:

Standard potential of the cell is the difference between right half-cell potential, where the reduction occurs, and left half-cell potential, where the oxidation occurs. Let's find the potentials of the half-cells for each case and then, subtracting the one for the left half-cell from the one for the right half-cell, find the $E_{0}$ :

a. $\mathrm{Fe}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}, \phi_{0} = +0.77 \mathrm{~V}$

$$\mathrm {Cd} ^ {2 +} + 2 \mathrm {e} ^ {-} \rightarrow \mathrm {Cd} (\mathrm {s}), \phi_ {0} = - 0.40 \mathrm {V}$$ $$E _ {o} = E _ {r i g h t} - E _ {l e f t} = + 0. 7 7 - (- 0. 4 0) = + 1. 1 7 \mathrm {V}$$

b. $\mathrm{MnO}_2(\mathrm{s}) + 4\mathrm{H}^+ +2\mathrm{e}^-\rightarrow \mathrm{Mn}^{2 + } + 2\mathrm{H}_2\mathrm{O},\phi_0 = +1.23\mathrm{V}$

$$\mathrm {Ag} _ {2} \mathrm {CrO} _ {4} (\mathrm {s}) + 2 \mathrm {e} ^ {-} \rightarrow 2 \mathrm {Ag} (\mathrm {s}) + \mathrm {CrO} _ {4} ^ {2 -} (\mathrm {aq}), \phi_ {0} = + 0. 4 5 \mathrm {V}$$ $$E _ {o} = E _ {r i g h t} - E _ {l e f t} = + 1. 2 3 - 0. 4 5 = 0. 7 8 \mathrm {V}$$

Balanced cell reaction equations are:

a. $2\mathrm{Fe}^{3+} + \mathrm{Cd(s)} \rightarrow 2\mathrm{Fe}^{2+} + \mathrm{Cd}^{2+}$ , $\mathrm{Fe}^{3+}$ is reduced to $\mathrm{Fe}^{2+}$ , $\mathrm{Cd(s)}$ is oxidized to $\mathrm{Cd}^{2+}$ .

b. $\mathrm{MnO}_2(\mathrm{s}) + 4\mathrm{H}^+ + 2\mathrm{e}^- + 2\mathrm{Ag}(\mathrm{s}) + \mathrm{CrO}_4^{2-}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+} + 2\mathrm{H}_2\mathrm{O} + \mathrm{Ag}_2\mathrm{CrO}_4(\mathrm{s}), \mathrm{Mn}$ is reduced from $+4$ to $+2$ , and $\mathrm{Ag}$ is oxidized from to $+1$ .

Now, let's calculate the reaction quotient:

a. $Q = \frac{c(Fe^{2+})^2 \cdot c(Cd^{2+})}{c(Fe^{3+})^2} = \frac{0.5^2 \cdot 1.24}{0.75^2} = 0.55$

b. $Q = \frac{c(Mn^{2+})}{c(CrO_4^{2-}) \cdot c(H^+)^4} = \frac{0.1}{2 \cdot 2^4} = 3.13 \cdot 10^{-3}$

And finally, we can calculate $E_{\text{cell}}$ :

a. $E_{\text{cell}} = E_o - \frac{RT}{nF} \ln Q = 1.17 - \frac{0.05916}{2} \cdot \lg(0.55) = 1.18 \, \text{V}$

b. $E_{\text{cell}} = E_o - \frac{RT}{nF} \ln Q = 0.78 - \frac{0.05916}{2} \cdot \lg(3.13 \cdot 10^{-3}) = 0.85 \, \text{V}$

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